phân loại và phương pháp giải đại số 10

NGUYEN ANH TRUdNG - NGUYEN PHU KHANH DAU THANH KY - NGUYEN MINH NHIEN NGUYEN TAN SIENG - DO NGOC THUY Nhdm gi£o viSn chuy§n todn THPT PHAN LOAI & PH V0N6 PHAP 6IAI Tai han On nb^t c

NGUYEN ANH TRUdNG - NGUYEN PHU KHANH

DAU THANH KY - NGUYEN MINH NHIEN

NGUYEN TAN SIENG - DO NGOC THUY

(Nhdm gi£o viSn chuy§n todn THPT)

PHAN LOAI &

PH V0N6 PHAP 6IAI

(Tai han On nb^t c6 cbinh If vk ho sung)

Danh cho hoc sinh Idp 10 on tap

va nang cao kien thufc

Bien soan theo noi dung, chifofng trinh

sach gi^o khoa Bo GD & DT

Cac em hoc sinh than men!

' T h a n loai va p h u o n g phap giai D a i so 10" la mot trong n h u n g cuon

thuoc bo sach "Phdn loai vd phuong phdp gidi: lap 10, 11, 12 ", do n h o m tac

gia chuyen toan THPT bien soan

V 6 i each viet khoa hoc va sinh dong giup ban doc tiep can v6i mon Toan

mpt each t u nhien, khong ap luc, ban doc tro nen t u tin va nang dong hon;

hieu ro ban chat, biet each phan tich de t i m ra trong tam cua van de va biet giai

thich, lap luan cho t u n g bai toan Sir da dang ciia he thong bai tap va tinh

Huong giup ban doc iuon hiVng thu khi giai toan

Tac gia chii trong bien soan nhCing cau hoi mo, no! d u n g co ban bam sat

sach giao khoa va cau true de thi dai hoe, dong thoi phan bai tap thanh cac

dang toan co loi giai chi tiet Hien nay de thi dai hoc khong kho, to hop cua

nhieu van de d o n gian, n h u n g chiia nhieu cau hoi m o ne'u khong nam chae ly

thuyet se lung tiing trong viee tim loi giai bai toan Vol mot bai toan, khong

nen thoa man ngay voi mot loi giai m i n h vua tim duoc ma phai co gang tim

nhieu each giai nha't cho bai toan do, moi mot each giai se co them phan kien

thuc moi on tap

M o n Toan la mot m o n rat ua phong each tai t u , n h u n g phai la tai t u mot

each sang tao va thong m i n h Khi giai mot bai toan, thay vi diing thoi gian de

luc Ipi t r i nho, thi ta can phai suy nghl phan tich de tim ra p h u o n g phap giai

quyet bai toan do Doi voi Toan hoc, khong co trang sach nao la thua T u n g

trang^ t u n g dong deu phai hieu M o n Toan doi hoi phai kien nhan va ben bi

ngay t u nhCrng bai tap don gian nha't, nhirng kien thuc ca ban nha't V i chinh

nh&ng kien thuc co ban m o i giup ban doc hieu duoc nhi>ng kien thuc nang cao

sau nay L u d w i g Van Beethoven: "Giot nuoc co the lam mon tang da, khong

phai v i giot nude co sue manh, ma do nuoc ehay lien tuc ngay dem Chi co su

phan da'u khong met moi m o i dem lai tai nang Do do ta eo the khSng dinh,

khong nhich t u n g bude thi khong bao gid cd the'di xa ngan d a m "

Mae d i i tac gia da danh nhieu tam huye't cho cudn sach, song su sai sdt la

dieu khd tranh khdi Chung toi rat mong nhan duoc su phan bien va gdp y quy

bau eiia quy doe gia de nhu-ng Ian tai ban sau eud'n sach duoc hoan thien hon

Thay mat n h d m bien soan

N g u y e n Phu Khanh

Cty TNHH MTV DWH Khang Vi.'(

l^huon^ 1 MENH D E - TAP HOOP

§1 MENH DE VA MENH DE CHUA BIEN

A TOM T A T L Y T H U Y E T

<i)inhnghia: ^ r;,^ ; ,! ; '

Mln/j (fe la mot eau khang dinh Diin^hoac S f l i ,, ,-, ^

M o t menh de khong the vua diing hoae vua sai ^ r

2 M$nh de phu dinh:

Cho menh de P M e n h de " K h o n g phai P " goi la menh dephu dinh cua P

K y hieu la P Ne'u P d u n g thi P sai, ne'u P sai thi P d u n g

3 Menh de keo theo vd menh de ddo

Cho hai menh de P va Q NJenh de "ne'u P thi Q" goi la menh dekeo theo

K y hieu la P => Q M e n h de P => Q chi sai khi P d u n g Q sai

Cho menh de P => Q K h i d d menh de Q => P goi la menh de ddo eua

5 Menh de chiia bien

M e n h de ehiia bien la mot cau khMng djnh chua bien nhan gia trj trong mpt tap X nao d d ma v d i moi gia t n cua bien thuoc X ta duoc mgt m^nh de

V i d y : P ( n ) : "n chia het cho 5 " v d i n la sd'tu nhien

P{x; y ) : " 2x + y = 5 " vdi X, y la so thuc

6 Cac ki hi^u V , 3 menh de phu dinh cua menh de cd chua ki hieu V ,3

K i hieu V: doc la vdi moi, 3: doc la ton tai

Phii d j n h cua m ^ n h d e " V x e X , P ( x ) " la menh de " 3x e X , P ( x ) "

Phu d j n h ciia menh d e " 3 x € X , P ( x ) " la menh de " Vx e X,P(x) " ^ ,

Phdn loai va phuamg phdpgiai Dai so'10

B CAC DANG TOAN VA PHl/QNG PHAP G I A I

| D A N G T O A N l XAC DINH MENH DE VA TJNHDUNG SAI CIJA MENHDE

Ca 1 C A C Vf D U MINH H O A

V i d \ 1: Cac cau sau day, cau nao la menh de, cau nao khong phai la menh

de? Ne'u la menh de hay cho biet menh de do d i i n g hay sai

(1) O day dep qua!

(2) Phuong trinh - 3x + 1 = 0 v6 nghiem

(3) 16 khong la so nguyen to

(4; ai p h u o n g trinh - 4x + 3 = 0 va x^ - Vx + 3 + 1 = 0 c6 nghiem chung

(5) So 7t CO Ion hon 3 hay khong?

(6) Italia v6 dich W o r l d c u p 2006

(7) H a i tam giac bang nhau k h i va chi khi chiing c6 dien tich bang nhau

(8) M o t t u giac la hinh thoi khi va chi khi no c6 hai d u o n g cheo vuong goc

v 6 i nhau

Lai gidi

Cau (1) va (5) khong la menh de(vi la cau cam than, cau hoi)

Cac cau (3), (4), (6), (8) la n h u n g menh de d u n g

Cau (2) va (7) la n h i f n g menh de sai

V i d u 2: Cho ba menh de sau, v o i n la so t u nhien

(1) n + 8 la so chinh p h u o n g

(2) C h u so tan ciing ciia n la 4

(3) n - l la so chinh p h u o n g

Biet rang c6 hai menh de diing va mot menh de sai Hay xac d j n h menh

de nao d u n g , menh de nao sai

Lcn gidi

Ta CO so chinh p h u o n g c6 cac dm so tan cung laO, 1, 4, 5, 6, 9 V i vay

- N h a n thay giua menh de (1) va (2) c6 mau thuan Boi v i , gia sir 2 menh de

nay dong thai la d u n g thi n + 8 c6 chir so tan ciing la 2 nen khong the la so

chinh p h u o n g Vay trong hai menh de nay phai c6 mot m f nh de la d i i n g va

mpt m^nh de la sai

- T u o n g t u , nhan thay giiia menh de (2) va (3) cung c6 mau thuan Boi v i , gia

sir m?nh de nay dong thoi la diing thi n - 1 c6 ch\x so tan cung la 3 nen

khong the la so chinh p h u o n g

Vay trong ba menh de tren thi menh de (1) va (3) la diing, con menh de (2)

la sai

ga 2 B A I T A P L U Y | N T A P

Bai 1.0: Cac cau sau day, cau nao la menh de, cau nao khong phai la m^nh de?

Ne'u la menh de hay cho biet menh de do d u n g hay sai

a) K h o n g di\xgc d i loi nay! , ? '•

b) Bay gio la may gio? ; n : | r ' c < H o y

c) Chieh tranh the gioi Ian t h i i hai ket thiic nam 1 9 4 6 I r „:; v ! =\

d) 1 6 c h i a 3 d u l e) 2003 khong la so nguyen to ,

£) v'5 la so v6 t i g) H a i d u o n g tron phan biet c6 nhieu nhat la hai diem chung ^

Huang dan gidi

Cau khong phai m^nh de la a), b) d,^ Cau d), f) la menh de dung Cau e) sai Cau g) d i i n g

-Bai 1.1: Tai Tiger C u p 98 c6 bon doi lot vao vong ban ket: Vi^t N a m , Singapor, Thai Lan va Indonexia Truoc k h i thi dau vong ban ket, ba ban Dung, Quang, T r u n g d u doan n h u sau:

Dung: Singapor n h i , con Thai Lan ba

Trung: Singapor nhat va Indonexia nhi

Ket qua, m o i ban d u doan d u n g mot doi va sai mpt d o i H o i m o i d p i da dat giai may?

Huang dan gidi ,,

-Ta xet d u doan ciia ban D u n g + Ne'u Singgapor n h i thi Singapor nhat la sai do do Indonexia nhi la dung(mau thuan)

+ N h u vay Thai lan t h i i ba la d u n g suy ra Vi?t N a m nhi, Singapor nhat va Indonexia t h i i t u

D A N G T O A N 2 : CAC PHEP TOAN VE MENH DE

Cac phep todn menh deduce sit dung nham muc dich ket not cac menh delai vox nhau tao ra mot menh de moi Mot so cac phep todn menh de la : Menh de phu dinhiphep phu djnh), menh dekeo theoiphep keo theo), menh deddo, menh detitang duongiphep tuvngducmg)

Phan lo^i vA phucmg phdp gidi Dai sd'lO

Ta CO cac m^nh de phii dinh la

P : " H a i d u o n g cheo ciia hinh thoi khong vuong goc v o i nhau", menh de

nay sai

Q : "6 khong phai la so'nguyen to", menh de nay d u n g

R : "Tong hai canh ciia mot tarn giac nho han hoac bang canh con lai",

m f nh de nay sai

S: " 5 < - 3 " , menh de nay sai

K : " p h u a n g trinh - 2x^ + 2 = 0 v6 nghi^m", m|nh de nay d u n g v i

x'^-2x^ +2 = (x^ if +l>0

H : " ( V 3 - V i 2 ) = 3 ", m i n h d e nay sai

V i dv 2: Phat bieu m^nh de P => Q va phat bieu m^nh de dao, xet tinh d i i n g

sai ciia no

a) P: " T u giac A B C D la hinh thoi" va Q : " T u giac ABCD, A C va BD cat nhau

tai trung diem m o i d u o n g "

b) P: " 2 > 9 " va Q : " 4 < 3 "

c) P: " T a m giac ABC vuong can tai A " va Q : " Tam giac ABC c6 A = 2B "

d) P: "Ngay 2 thang 9 la ngay Quoc Khanh ciia nuac Viet N a m " va Q: "Ngay

27 thang 7 la ngay thuong binh l i f t si"

Lai gidi

a) M e n h de P Q la "Neu t u giac A B C D la hinh thoi thi A C va BD cat nhau

tai trung diem moi duang", menh de nay dung

M ? n h de dao la Q ri> P: "Neu t u giac A B C D c6 A C va BD c^t nhau tai trung

diem m o i d u a n g thi A B C D la hinh thoi", menh de nay sai

b) M e n h de P => Q la "Neu 2 > 9 thi 4 < 3", menh de nay d i i n g v i menh de P sai Menh de dao la Q => P: "Neu 4 < 3 thi 2 > 9", menh de nay diing vi m^nh de

c) M e n h de P => Q la "Neu tam giac ABC vuong can tai A thi A = 2B ", menh

de nay d i i n g Menh de dao la Q => P: "Neu tam giac ABC c6 A = 2B thi no vuong can tai A", menh de nay sai

d) Menh de P => Q la "Neu ngay 2 thang 9 la ngay Quoc Khanh ciia nuoc Viet

N a m thi ngay 27 thang 7 la ngay thuong binh liet sT" ' ' Menh de dao la Q => P: "Neu ngay 27 thang 7 la ngay t h u o n g binh liet sT thi ngay 2 thang 9 la ngay Quoc Khanh ciia nuoc Viet N a m "

Hai menh de tren deu diing vi menh de P, Q deu diing

V i d u 3: Phat bieu menh de P o Q bang hai each va xet ti'nh diing sai ciia no

a) P: " T i i giac A B C D la hinh thoi" va Q: " T u giac A B C D la hinh binh hanh

CO hai d u o n g cheo vuong goc voi nhau"

b) P: "Ba't p h u o n g trinh Vx^ - 3 x > 1 c6 nghiem" va Q: " ^(-1)^ - 3 ( - l ) > 1 "

Lai gidi

a) Ta CO menh de P <=> Q diing vi menh de P => Q, Q => P deu d i i n g va duoc

phat bieu bang hai each n h u sau:

" T i i giac A B C D la hinh thoi k h i va chi khi t i i giac A B C D la hinh binh hanh

CO hai d u o n g cheo vuong goc voi nhau" va

" T u giac A B C D la hinh thoi neu va chi neu t i i giac A B C D la hinh binh hanh

CO hai d u a n g cheo vuong goc voi nhau"

b) Ta CO menh de P <=> Q diing vi menh de P, Q deu dung(do do m^nh de P => Q,

Q => P deu diing) va dugc phat bieu bSng hai each n h u sau:

"Bat p h u o n g trinh \/x^ - 3 x > 1 c6 nghiem khi va chi k h i -3.(-\) > 1 "

va "Bat phuong trinh - 3 x >1 c6 nghiem neu va chi neu ^(-1)^ - 3 ( - l ) > 1 "

ea 2 BAI TAP LUYIN TAP

Bai 1.2: Neu menh de phu dinh ciia cac menh de sau, cho bie't menh de nay

d i i n g hay sai?

P: "Trong tam giac tong ba goc bang 180""

Q:"(V3-V27)^ la so n g u y e n "

R: "Viet N a m v6 djch W o r l d c u p 2020" ' '

Phdn loai va phuong phdpgidi Dai so 10

S : " - ^ > - 2 "

2

K: "Bat p h u o n g trinh x ^ ° " > 2030 v6 nghiem "

Ta CO cac menh de phu dinh la

P : " Trong tam giac tong ba goc khong bang 180°", menh de nay sai

Q : " (N/3 - \/27) khong phai la so nguyen ", menh de nay sai

R : " Viet N a m khong v6 djch Worldcup 2020", menh de nay chua xac djnh

dirge d i i n g hay sai

S: " < - 2 " , menh de nay d u n g

K :" Bat p h u o n g trinh x^"^'^ > 2030 c6 nghiem ", menh de nay d u n g

Bai 1.3: Phat bieu menh de P => Q va phat bieu menh de dao, xet tinh diing sai

ciia no

a) P: " T u giac A B C D la hinh chii nhat" va Q: " T u giac A B C D c6 hai d u o n g

thang A C va BD vuong goc v6i nhau"

b) P: "-V3>-V2" v a Q : "{-^fsf > {-^f "

c) P: "Tam giac ABC c6 A = B + C " va Q: "Tam giac ABC c6 BC^ = AB^ + AC^ "

d) P: "To H i r u la nha Toan hoc Ion cua Viet N a m " va Q: "Evariste Galois la nhk

Tha loi lac cua T h e ' g i o i "

Huang dan gidi

a) P => Q : " N e u t u giac A B C D la hinh c h u nhat thi t u giac A B C D c6 hai

d u o n g thang A C va BD vuong goc voi nhau", menh de sai

Q => P: " N e u t u giac A B C D hai d u o n g thang A C va BD vuong goc v o l

nhau thi t u giac A B C D c6 la hinh chCr nhat", m?nh de sai

b) P ^ Q : " N e u -73 > -y/2 thi [-^f > [-^f ", menh de d u n g

P ^ Q : " N e u (-^f > (-^flf thi - > -72 ", menh de sai

c) P=>Q: "Neu tam giac ABC c6 A = B + C thi tam giac ABC c6 BC?=AB?+AC?"

Q => P: "Neu tam giac ABC c6 BC^ = AB^ + AC^ thi A = B + C "

Ca hai menh de deu diing

d) P^Q:" Neu To H u u la nha Toan hoc Ion cua Viet N a m thi Evariste Galois

la nha T h o loi lac cua The'gioi", Q => P :" Neu Evariste Galois la nha Tho loi

lac ciia The'gioi thi To H i r u la nha Toan hpc Ion ciia Viet N a m " H a i m^nh

de diing

Cty TNHHMTV nvv)l Khang Vict

Bai 1.4: Phat bieu m^nh de P <=> Q bang hai each va va xet tinh d i i n g sai ciia no a) Cho t u giac ABDC Xet hai menh de | P: "Tu giac A B C D la hinh vuong"

Q: " T i i giac A B C D la hinh chu nhat c6 hai d u o n g cheo v u o n g goc voi nhau"

b) P: "Bat p h u o n g trinh x^ - 3 x + l > 0 c6 nghiem" va Q: " Bat p h u o n g trinh x^ - 3x + 1 < 0 v6 nghiem"

Huang dan gidi

a) Ta CO menh de P <=> Q diing vi menh de P Q, Q => P deu d i i n g va dugc

phat bieu bang hai each n h u sau:

" T u giac A B C D la hinh vuong khi va chi k h i t i i giac A B C D la hinh chu

nhat CO hai d u o n g eheo bang vuong goc v o i nhau " va

" T u giac A B C D la hinh vuong neu va chi neu t u giac A B C D la hinh chii

nhat CO hai d u o n g cheo vuong goc voi nhau "

b) Ta CO menh de P o Q sai v i menh de P d i i n g con Q sai

Phat bieu menh de P <=> Q b3ng hai each

"Bat p h u o n g trinh x^ - 3x + 1 > 0 c6 nghiem khi va chi khi ba't p h u o n g trinh x^ - 3x + 1 < 0 v6 nghiem" va "Bat phuong trinh x^ - 3x + 1 > 0 c6 nghiem neu va chi neu ba't p h u o n g trinh x^ - 3x + 1 < 0 v6 nghiem"

a-\/3

A : " N e u AABC deu c6 canh bang a, d u o n g cao la h thi h = " ;

B : " T i i giac CO bon canh bSng nhau la hinh v u o n g " ;

Huang dan gidi

Ta CO A va D la cac menh de diing, B va C la cac m^nh de sai D o do ;

a) M e n h de A = > B sai v i A diing va B sai ' >

M e n h de A D diing vi A va D deu diing ,

M e n h de B =:> C d i i n g v i B sai

b) M e n h de A B sai v i menh de A B sai (Hoac A diing va B sai) M e n h de

B <=> C d i i n g v i hai menh de B va C deu sai

M e n h de A Ci> D diing v i hai m^nh de A va D deu diing

Phiht loai va pltucnig phiip giiii Dai so'10

Bai 1.6: Hay phat bieu menh de keo theo P => Q, Q => P va xet tinh d i i n g sai

cua menh de

nay-a) Cho tiV giac A B C D va hai menh de: ' ' •

P: "Tong 2 goc doi cua t u giac loi bang 180" " va Q: " T u giac noi tiep duoc

d u o n g tron "

b) P : " >/2 - > - 1 " va Q: "(^/2 - Vsf > ( - 1 ) ' "

Htt&ttg dan gidi

a) P => Q : " Neu tong 2 goc doi cua tiV giac loi bang 180" thi tiV giac do npi tiep

du(?c d u o n g tron "

Q P: "Neu t u giac khong noi tiep d u o n g tron thi tong 2 goc doi cua t u

giac do bang 180""

M e n h de P => Q dung, menh de Q => P sai

b) P Q : " Neu V2 - V3 > - 1 thi (V2 - S f > {-if "

ffll.CAC Vl DU MINH HOA

V i du 1: Cho m^nh de chua bien " P ( x ) : x > x"'", xet tinh d u n g sai ciia cac

menh de sau:

a) P O )

c) V x e N , P ( x )

b) P d) 3 x e N , P(x)

3 day la menh de diing

c) Ta CO Vx € N , X > x'"* la m^nh de sai vi P ( l ) la menh de sai

d) Ta CO 3x € N , x > x'' la m f n h de diing v i x - x"' = x ( l - x ) ( l + x) < 0 v o i moi

so' t y nhien

Cty TNHH MTV DWH Khang Viet

V i du 2: D u n g cac ki hieu de vie't cac cau sau va vie't menh de phu djnh ciia no

a) Tich ciia ba so'tu nhien lien tiep chia he't cho sau b) V o i m p i so thuc binh phuong ciia la mot so khong am i c) Co mot so nguyen ma binh phuong ciia no bang chi'nh no

d) Co mot so hCru ti ma nghich dao ciia no Ion hon chinh no

Loi gidi i - u i j \

a) Ta CO P: Vn e N , n ( n + l ) ( n + 2):6, menh de phii djnh la r

P : 3 n e N , n ( n + l ) ( n + 2)/6

^'/^.-b) Ta CO Q : Vx e M, x^ > 0, menh de phii djnh la Q :3x 6 K, x^ < 0 c) Ta CO R : 3n G Z, n^ = n , menh de phii djnh ia R : Vn e Z, n^ ;t n ' ' '

d) 3q € Q, — > q , menh de phu dinh la V q e Q, i < q

q M

V i du 3: Xac d j n h tinh diing sai ciia menh de sau va t i m phii dinh ciia no :

a) A: " V x e R , x^ > 0 "

b) B: " Ton tai so t u nhien deu la so nguyen to"

c) C: " 3 x 6 N , X chia he't cho x + 1 "

d) D : " V n £ N , n ' ^ - n ^ + l la hop so"

e) E: "Ton tai hinh thang la hinh vuong "

f) F: "Ton tai so thuc a sao cho a + 1 + — < 2 "

a + 1

Lai gidi

a) Menh de A diing va A : 3x € R, x ' < 0 b) Menh de B diing va B : "Voi moi s o t u nhien deu khong phai la so nguyen to"

c) M e n h d e C sai va C : " Vx e N , x / ( x + 1)"

d) M e n h de D sai vi voi n = 2 ta c6 n'* - n^ +1 = 13 khong phai la hop so

M ^ n h de phii d j n h la D : " 3n e N , n"* - n^ +1 la so so nguyen to"

e) Menh de E d i i n g va E : " V o i m p i hinh thang deu khong la hinh vuong "

0 M e n h de F diing va menh de phii djnh la F: "Voi m o i so thuc a thi

a + l + > 2

a + 1

Phdtt loai va phucnig phdp gM Dai so 10

c) M e n h de 3x G N , n^ + 3 chia het cho 4 diing v i n = 1 e N va n^ + 3 = 4:4

M^nh de phii d i n h la " Vx e N , n + 3 khong chia het cho 4"

d) M e n de 3q e Q, 2q^ - 1 = 0 sai Menh de phu dinh la Vq e Q, 2q^ - 1 0

e) M e n h de " 3 n G N , n ( n + l ) la mot so chinh p h u o n g " diing M | n h de phii

d i n h la " Vn e N , n (n +1) khong phai la mot so chinh p h u o n g "

Bai 1.8:

a) V o i n e N , cho menh de chua bien P(n):"n^ + 2 chia het cho 4" Xet tinh

diing sai ciia menh de P(2007)

b) Xet tinh diing sai cua menh de P{n) : " 3n e N * , ^ n ( n +1) chia het cho 11"

^ Huang dan gidi

a) Ta CO : V o i n = 2007 thi n^ + 2 = 2007^ + 2 la so le nen khong chia het cho 4

Vay P(2007) la menh de sai

b) Xet bieu thiVc ^ , voi n G N * ta c6 :

2 Voi n = 10 thi = 55 : chia het cho 11 Vay menh de da cho la menh

de diing |

Cty TNHH MTV DWH Khang Vjc-i

Bai 1.9:

a) Cho menh de P : "Voi moi so thuc x, neu x la so h i i u ti thi 2x la sd'hiru t i "

D u n g k i hieu viet P, P va xac djnh tinh diing - sai ciia no

b) Phat bieu M D dao ciia P va chiing to M D do la diing Phat bieu M D d u o i dang M D tuong d u o n g

Huang dan gidi ' f i ' sr

a) Menh d e P " V x G R , X G Q = > 2 X G Q " M D diing

P: "3x G R,x 6 Q 2x g Q " M D sai b) M D dao ciia P la " V o i moi so thuc x, X G Q k h i va chi k h i 2 X G Q " Hay

c) Hay phat bieu m^nh de " V n e N , A(n) <r> B ( n ) "

Huang dan gidi

a) A(n) => B(n) : " N e u n la so chan thi n^ la so chan" Day la menh de diing, v i khi do n = 2k (k G N ) => n^ = 4k2 la so chan

b) " Vn G N , B(n) => A ( n ) " : V o i moi so t u nhien n, neu n^ la so chan thi n la so chan

c) " Vn € N , A ( n ) <=> B(n) " : V o i moi so t u nhien n, n la so c h i n k h i va chi k h i n^

la sochan

Bai 1,11: Xet tinh d i i n g sai ciia cac m^nh de sau:

a) P : " V \ G R , V y G R : x + y = l " b) Q :"3x G R,3y G R : x + y = 2"

c) R : " T G R , V y G R : x + y = 3 " d) S :" Vx G R,3y G R : x + y = 4 "

Huang dan gidi ' '

a) M e n h de P sai vi ch^ng han X = 1GIR, y = 2 G K n h u n g x + y ^ 1 b) M e n h de Q d i i n g v i X = y = 1 => x + y = 2 •>

c) V i X + y = 3 nen v o i m o i y G M thi luon ton tai x = 3 - y do do m f nh de R

d i i n g ' d) M e n h de S diing "

Phatt loai va fhucntg phdp giai D(it so

§2: AP DUNG MENH D E VAO SUY LUAN TOAN HOC

A TOM T A T L Y T H U Y E T

/ 't)inh li va chung minh dinh U

• Trong toan hoc djnh ly la mot menh de diing Nhieu dinh ly dugc phat bieu

d u o i d a n g " V x e X , P(x)=:>Q(x)", P ( X ) , Q ( X ) la cac menh de chiia bien

• Co hai each de chiing minh dinh li d u o i dang tren

Cach 1: Chung m i n h true tie'p gom cac budc sau:

- Lay x e X b a ' t k y m a P ( x ) d u n g

- Chung m i n h Q ( x ) dung(bang suy luan va kien thuc toan hoc da biet)

Cach 2: Chung m i n h bang phan djnh li gom cac buoc sau:

- Gia su ton tai X Q e X sao cho P ( x „ ) d u n g va Q ( X ( , ) sai

- DiJng suy luan va cac kien thuc toan hoc de di de'n mau thuan

2 D j n h li dao, dieu ki?n can, dieu ki?n dii, dieu ki^n can va du

• Cho djnh li d u o i dang "Vx 6 X, P ( x ) =:> Q ( x ) " (1) K h i do

P( x ) la dieti kien dii deed Q ( x )

Q( x ) \a dieu kien can de CO P ( x )

• Menh de V x € X , Q ( x ) = > P ( x ) dung thi duoc goi dinh li dao ciia djnh li

dang (1)

Liic do (1) duoc goi la dinh Ixj thuan va khi do c6 the gop lai thanh mot djnh li

Vx G X, Q ( x ) o P ( x ) , ta goi la " P ( x ) la dieu kien can va dii de c6 Q ( x ) "

Ngoai ra con n o i " P ( x ) ne'u va chi neu Q ( x ) " , " P ( x ) khi va chi khi Q ( x ) " ,

B CAC DANG TOAN VA PHl/QNG PHAP G I A I

D A N G T O A N 1: PHUONG PHAP CHUNG MINH BANG PHAN CHUNG

Q L C A C V ( D U M I N H H O A

V i du 1: C h u n g m i n h rSng voi moi so t u nhien n, n ' ' chia het cho 3 thi n

chia he't cho 3

Lai giai

Gia su n khong chia he't cho 3 khi do n = 3k + 1 hoac n = 3k + 2 , k e Z

; V o i n = 3k + 1 ta c6 n^ = (3k + i f = 27k^ + 27k^ + 9k +1 khong chia he't cho

3 (mau thuan)

Voi n = 3k + 2 ta c6 n ^ = ( 3 k + 2)^ = 2 7 k ^ + 5 4 k ^ + 3 6 k + 4 khong chia het

cho ba (mau thuan)

Vay n chia he't cho 3

V i d u 2: Cho tarn thuc f (x) = ax^ + bx + c, a 0 Chung minh rang ne'u ton tai

so thuc a sao cho a.f ( a ) < 0 thi phuong trinh f (x) = 0 luon c6 nghiem

Khi do ta co: af(x) = a' ^ b ^ 2 X + •

2a

\ itiv.tn

- - > 0 , V x e R

4 * , » v w i '

Suy ra khong ton tai a de af ( a ) < 0 , trai voi gia thiet

Vay dieu ta gia su 6 tren la sai, hay phuong trinh da cho luon c6 nghiem

V i du 3: C h u n g minh rang mot tarn giac c6 duong trung tuye'n vua la phan

giac xua't phat t u mot dinh la tam giac can tai dinh do

Do do A L = L D hay L la trung diem cua BD B

Suy ra L H la d u o n g trung binh cua tam giac CBD

=> L H / / D C dieu nay mau thuan vi L H , D C Ccit nhau tai A Vay tam giac ABC can tai A ;

Q 2 B A I T A P L U Y E N T A P Bai 1.12: Chung m i n h b^lng phuong phap phan chung: Ne'u p h u o n g trinh bac

hai ax^ + bx + c = 0 v6 nghiem thi a va c cung da'u

Huong dan gidi

Gia su p h u o n g trinh v6 nghiem va a, c trai da'u Voi dieu ki^n a, c trai da'u

CO a.c<0 suy ra A = b^ - 4ac = b^ + 4(-ac) > 0

Nen p h u o n g trinh co hai nghiem phan biet, dieu nay mau thuan v o i gia thiet p h u o n g trinh vo nghiem

Vay p h u o n g trinh v6 nghiem thi a, c phai cung da'u

Bai 1.13: Chung minh bang phuong phap phan chung: Ne'u hai so'nguyen duong

CO tong binh phuong chia he't cho 3 thi ca hai so do phai chia he't cho 3

Phdn lo^i va phuang phdp gi&i Dai so 10

Huang dan gidi

Gia su trong hai so' nguyen d u a n g a va b c6 it nhat m o t so khong chia het

cho 3, chang han a khong chia het cho 3 The thi a c6 dang: a = 3k + 1 hoac

a = 3k+2 Luc do a^ = 3m + 1 , nen neu b chia het cho 3 hoac b khong chia het

cho 3 thi a^ + b^ cung co dang: 3n + 1 hoac 3n + 2, tuc la a^ + b^ khong chia

het cho 3, trai gia thie't Vay neu a^ + b^ chia het cho 3 t h i ca a va b deu chia

het cho 3

Bai 1.14: C h u n g m i n h rang : Neu do dai cac canh cua tam giac thoa man bat

d i n g thuc a^ + b^ > 5c^ thi c la do dai canh nho nhat cua tam giac

Huong dan gidi

Gia su c khong phai la canh nho nhat cua tam giac

K h o n g mat tinh tong quat, gia sua < c => a^ < c^ (1)

Theo bat d i n g thuc trong tam giac, ta co b < a + c b^ < (a + c)^ (2)

D o a < c ^ ( a + c)^ <4c2 (3)

T u (2) va (3) suy ra b^ < 4c^ (4)

Cong ve v o i ve'(l) va (4) ta co a^ + b^ < 5c^ mau thuan v o i gia thie't

Vay c la canh nho nhat cua tam giac

Bai 1.15; Cho a, b, c d u o n g nho hon I C h u n g m i n h rang it nhat mot trong ba

ba't d i n g thuc sau sai a ( l - b ) > i , b ( l - c) > c ( l - a) > i

Huong dan gidi

Gia sir ca ba bat d i n g thuc deu dung

Khi do, nhan theo ve'ciia cac bat d i n g thuc tren ta duoc:

Vay CO ft nhat m o t trong cac bat dang thuc da cho la sai (dpcm)

Bai 1.16: Neu a ] a 2 > 2 ( b 5 + b2) thi it nhat mot trong hai p h u a n g trinh + ajx + bj = 0, + a2X + b2 = 0 co nghiem

Huong dan gia i

Gia su ca hai p h u a n g trinh tren v6 nghiem Khi do D , = a i ^ - 4 b i < 0, D ] = - 4b2 < 0

_ ^ a / - 4 b i + a 2 ^ - 4 b 2 <Oc:>a,^+32^ < 4 ( b i + b 2 ) (1) '

M a (a, - a2 f > 0 <=> a? + a^ > 23^32 (2)

T u (1) va (2) suy ra 23^32 < 4 ( b j + b2) hay 3^32 < 2{b] + b j ) trai gia thie't

Vay phai co it nhat 1 trong hai so A i , A2 Ion han 0 do do it nhst 1 trong 2

p h u a n g trinh x + ajx + b^ = 0 , x + a2X + b2 = 0 co nghiem

Bai 1.17: C h u n g m i n h rang sfl la so v6 t i

Huong dan gidi ,^

De dang c h u n g m i n h dugc neu n^ l a s o c h S n t h i n laso'chan

Gia sir V2 la so hCru t i , tuc la J2= — , trong do m , n e N*, ( m , n) = 1

n

Tit N/2 = — => m^ = 2n^ => la so chan

n

=> m la so chan => m = 2k, k e N Tir m^ = 2n^ =:> 4k^ = 2n^ :=> - 2k^ => n^ la so c h i n => n la so chan

Do do m chan, n chan, mau thuan vai ( m , n) = 1 Vay N/2 la so v6 ti

a + b + o O (1)

Bai 1.18: Cho cac so a, b, c thoa cac dieu ki§n : • ab + be + ca > 0 (2)

abc > 0 (3) Chirng m i n h rang ca ba so a, b, c deu d u o n g

Huong dan gidi

Gia sir ba so a, b, c khong dong thai la so d u a n g Vay co it nhat mot so khong d u a n g

Do a, b, c CO vai tro binh d i n g nen ta co the gia sir: a < 0

Neu a = 0 thi mau thuan v o i (3) Ne'u a < 0 thi tir (3) => be < 0

Ta CO (2) <=> a(b + c) > -be => 3(b

^ b + c < 0 a + b + c < 0 mauth|ulli%y»P TiWH BiNH THUAfJ

Vav ca b 3 so a, b, c deu duone

Than loai va phumtg phapgiai Dai so 10

Bai 1.19: Chung m i n h bang phan chiing dinh li sau : " N e u tarn giac ABC c6 cac

d u a n g phan giac trong BE, CF bang nhau, thi tarn giac ABC can"

Huang dan gidi

• Neu B > C thi ta dung hinh binh hanh BEDF n h u hinh ve

Xet cac tarn giac BCE va CBF, ta thay : ^

BC Chung, BE = CF, BF > CE nen Q > B^ =^ C > B M a u thuan

• Truong hop C > B, chung m i n h hoan toan tuong t u n h u tren

Do do B = C Vay tarn giac ABC can tai A

Bai 1.20: Cho 7 doan thSng c6 do dai Ion hon 10 va nho hon 100 Chung m i n h

rang luon t i m dugc 3 doan de c6 the ghep thanh mgt tarn giac

Huang dan gidi

Truoc het sap xep cac doan da cho theo thu t u tang dan cua do dai a j , ,87

va chung minh rang trong day da xep luon tim dugc 3 doan lien tiep sao cho

tong cua 2 doan dau Ion hon doan cuoi (vi dieu kien de 3 doan c6 the ghep

thanh mgt tarn giac la tong cua 2 doan Ion hon doan thu ba)

Gia su dieu can chung m i n h la khong xay ra, nghia la dong thoi xay ra cac

bat dang thuc sau: aj <a3; a2 +33 <a^; ; +a(,<ay

T u gia thie't a^, aj c6 gia tri Ian hon 10, ta nhan dugc 83 > 20 T u 82 > 10

va 83 > 20 ta nhan dugc a^ > 30, 35 > 50, ag > 80 va ay > 130 Dieu

ay > 130 la mau thuan voi gia thiet cac do dai nho hon 100 Co mau thuan

nay la do gia sir dieu can chung m i n h khong xay ra

Vay, luon ton tai 3 doan lien tiep sao cho tong cua 2 doan dau Ion hon doan

cuoi Hay noi each khac la 3 doan nay c6 the ghep thanh mgt tam giac

DANG TOAN 2: SUDUNG THUAT NGU DIEU KIEN CAN, BIEU KIEN

DU, DIEU KIEN CAN VADU ||

Ca 1 CAC Vf DU MINH HOA " * ' ' '

V i du 1: Cho djnh l i : "Cho so t u nhien n Neu n'^ chia het cho 5 thi n chia het

cho 5" D i n h l i nay dugc viet d u o i dang P => Q a) Hay xac d i n h cac menh de P va Q

b) Phat bieu dinh l i tren bang each dung thuat ngii "dieu kien can"

c) Phat bieu djnh l i tren b^ng each diing thuat ngii "dieu kien d u " d) Hay phat bieu djnh l i dao (neu c6) cua djnh l i tren roi dung cac thuat ngii

"dieu kien can va d u " phat bieu gop ca hai d i n h l i thuan va dao

Lai gidi

a) P : "n la so t u nhien va n^ chia het cho 5", Q : " n chia het cho 5"

b) Voi n la so t u nhien, n chia het cho 5 la dieu kien can de n'^ chia het cho 5 ; hoac phat bieu each khac : Voi n la so t u nhien, dieu kien can de chia het cho 5 la n chia het cho 5

c) Voi n la so t u nhien, n"^ chia het cho 5 la diiFu kien du de n chia het cho 5

d) D i n h l i dao : "Cho so t u nhien n, neu n chia het cho 5 thi n"^ chia het cho 5"

That vay, neu n = 5k thi n'^ = 5\k^: So nay chia het cho 5

Dieu kien can va du de n chia he't cho 5 la n'^ chia het cho 5

V i du 2: Phat bieu cac menh de sau voi thuat ngii "Dieu kien can", "Dieu kien d u " » f'

a) Neu hai tam giac bang nhau thi chung c6 dien tich bang nhau b) Neu so'nguyen d u o n g chia het cho 6 thi chia het cho 3 c) Neu hinh thang c6 hai duong cheo bang nhau thi no la hinh thang can d) Neu tam giac ABC vuong tai A va A H la duong cao thi AB^ = BC.BH

Lai gidi

a) Hai tam giac bang nhau la dieu kien du de chiing c6 dien tich bang nhau Hai tam giac c6 dien tich bSng nhau la dieu kien can de chiing bang nhau b) So'nguyen d u o n g chia het cho 6 la dieu kien d i i de no chia het cho 3

So nguyen d u o n g chia het cho 3 la dieu kien can de no chia het cho 6 c) H i n h thang c6 hai d u o n g cheo bang nhau la dieu k i | n d u de no la hinh thang can

H i n h thang can la dieu kien can de no c6 hai duong cheo bang nhau d) Tam giac ABC vuong tai A va A H la duong cao la dieu ki?n d i i de

A B ^ - B C B H ^.R

Tam giac ABC c6 AB^ = BC.BH la dieu i<ien can de no vuong tai A va A H

la d u o n g cao 1^ ' ^ '

Ca 2 BAI T A P L U Y E N T A P | •

Bai 1.21: Phat bieu cac djnh ly sau day bSng each sir d u n g khai niem " Dieu

kien can"," Dieu kien d u "

a) Neu trong mat phang, hai d u o n g thSng cijng vuong goc v o i d u o n g thang

t h i i 3 thi hai d u o n g thang do song song voi nhau

b) Neu so'nguyen d u o n g c6 chu tan cung bang 5 thi chia het cho 5

c) Neu t u giac la hinh thoi thi 2 d u o n g cheo vuong goc voi nhau

d) Neu 2 tam giac bang nhau thi chung c6 cac goc tuong u n g bang nhau

e) Neu so'nguyen d u o n g a chia het cho 24 thi chia het cho 4 va 6

i, Huang dan giai

a) Trong mat phang, hai d u o n g thang citing vuong goc v o i d u o n g thang t h u 3

la dieu kien d u de hai d u o n g thang do song song voi nhau

Trong mat phang, hai d u o n g thang do song song voi nhau la dieu kien can

de hai d u a n g thang cung vuong goc voi d u o n g thang t h i i 3

b) So'nguyen duong co chu so tan ci^mg bang 5 la dieu kien du de chia he't cho 5

So nguyen d u o n g chia he't cho 5 la dieu kien can de no co c\\\x so tan ciing

bang 5

c) T u giac la hinh thoi la dieu kien du de no co 2 duong cheo vuong goc voi nhau

T u giac CO hai d u o n g cheo vuong gck voi nhau la dieu kien can de no la

e) So'nguyen d u o n g a chia he't cho 24 Ih dieu kien du de no chia he't cho 4 va 6

So'nguyen duong a chia he't cho 4 va 6 la dieu k i f n can de no chia he't cho 24

Bai 1.22. D u n g thuat ngij' dieu kien can va ctu de phat bieu dinh l i sau

a) M o t tam giac la tam giac can, neu va chi neu no co hai goc bang nhau

b) T u giac la hinh binh hanh khi va chi khi t u giac co hai d u o n g cheo cat nhau

tai trung diem ciia m o i d u o n g

c) x > y o ^ > 3 / ^ • « ^ ^ - ^

d) T u giac M N P Q la hinh binh hanh khi va chi khi M N = QP ^ V * ^

Huong dan gidi

a) M p t tam giac la tam giac can la dieu kien can va d u de no co hai goc bang nhau

CtyTNHH MTV DWH KHaWg VieT

b) T u giac la hinh binh hanh la dieu kien can va du de t u giac co hai dirong

cheo cat nhau tai trung diem cua moi d u o n g g

c) X > y la dieu kien can va dude ;

d) Dieu kien can va du de t u giac M N P Q la hinh binh hanh la M N = QP

Bai 1.23: Su dung thuat ngCf "dieu kien can", "dieu kien d i i " de phat bieu dinh

li sau:

a) " N e u mot t u giac la hinh vuong thi no co bon canh bang nhau"

Co dinh l i dao cua djnh li tren khong , vi sao? M i i

b) " N e u mot t u giac la hinh thoi thi no co hai d u o n g cheo v u o n g goc"

Co djnh li dao cua djnh l i tren khong , vi sao? ^

Hu&ng dan gidi ^,

a) M o t t u giac la hinh vuong la dieu kien d u de no co 4 canh bang nhau

M o t t u giac CO 4 canh b3ng nhau la dieu kien can de no la hinh vuong

Khong CO djnh l i dao v i t u giac co 4 canh bang nhau co t h e l a hinh thoi

b) M o t t u giac la hinh thoi la dieu kien du de no co hai d u o n g cheo vuong goc

M o t t u giac CO hai d u o n g cheo vuong goc la dieu kien can de no la hinh thoi Khong CO d j n h l i dao vi t u giac co hai duong cheo vuong goc co the la hinh

vuong hoac mot da giac bat ki co hai duong cheo vuong goc

• Tap hop la mot khai niem co ban cua toan hoc, khong d j n h nghia

• Cach xac dinh tap hop:

-+ Liet ke cac phan tir: vie't cac phan tir cua tap hop trong hai dau moc + Chi ra tinh chat dac t r u n g cho cac phan t u cua tap hop

• Tap rong: la tap hop khong chua phan t u nao, k i hieu 0 ' \ ^ '

2 Tap hap con - Tap hap bdng nhau ^^J'"

• A c B o ( V x 6 A = ^ x e B ) ,^

Cac tinh chat:

+ A c A , V A + 0 c A , V A + A c B,B c C=> A c C

• A = B o ( A c B v a B c A ) o (Vx,^ ^ A « x £ B)

3 Mot so tap con cua tap hap sdthuc

Ten gQJ, ky hi^u Tap hgp Hinh bieu dien

Tap so' thyc''

Nua khoang (-co; a]

Nua khoang [a ;+co)

4 Cac phep todn tap hap

• Giao ciia hai tap hop: A n B o f x I x e A va xeB)

• Hop cua hai tap hop: A u B < = > | x l x 6 A hoac x e B)

• Hi^u cua hai tap hgp: A \ B < = > ( x l x 6 A va xiB}

Phan bii: Cho B c A thi C A B = A \

Tim tap hop X sao cho

a) Taco B\ = {-3;0;1}

Suy ra X c B \ thi cac tap hop X la

• 0 , {-3},{0}, {!}, {-3;0}, {-3;!}, {0;1}, {-3;0;1}

b) Taco {-4;-2;-l;2;3;4}cXc{-4;-3;-2;-l;0;l;2;3;4}suy r a t l p h g p X la {-4;-2;-1; 2; 3; 4}, {-4;-2;-3;-1; 2; 3; 4}, {-4;-2;-1;0;2; 3; 4}

ia» Huong dan gidi

Ta CO cac tap h g p A , B , C duoc viet d u o i dang neu cac tinh chat dac trung

la A = { x e N I | x | < 4 } , B = { x e N l x la so le nho han 10}, C = { n ^ l n l a s o t u

nhien nho h o n 6)

14

Bai 1.25 Cho tap h o p A = |x e M I ^ e Z

a) H a y xac d i n h tap A bang each liet ke cac phan t u

b) T i m tat ca cac tap con ciia tap hop A ''^

Huang dan gidi

Suy ra X c B \ thi cac tap hop X la

0 , { O } , | 3 } , { 4 } , { O ; 3 } , { O ; 4 } , { 3 ; 4 M O ; 3 ; 4 } b) Ta CO A \ = { - 2 ; - l } v6i X c6 d u n g hai phan t u k h i d o X = { - 2 ; - l }

Bai 1.27: Cho tap A = {-1;1;5;8}, B ="G6m cac uoc so nguyen d u o n g cua 16"

a) Viet tap A d u o i dang chi ra tinh chat dac trung ciia cac phan t u Viet tap B d u a l dang liet ke cac phan t u

b) Xac d i n h cac phep toan A n B, A u B, A \

Huong dan gidi

DANG TO AN 2: SUDUNG BIEU DO VEN DE GIAI TOAN

.>"X

a IK

'Phucmg phdp gidi

• Chuyen bai toan ve ngon ngi> tap hop

• Su* dung bieu do ven de minh hoa cac tap hop

• Dua vao bieu do ven ta thie't lap dugc dang thiic (hoac phuang trinh he

phuang trinh) t u do tim du'oc ke't qua bai toan ' i j i

Trong dang toan nay ta ki hieu n (X) la so phan t u cua tap X

£ • 1 C A C V l D U M I N H H O A ^ qv" q(>! n i l ••\ (t :.,) A i:>i V U r

Vi du 1: M o i hoc sinh cua lop lOAi deu biet choi da cau hoac cau long, bie't

rang c6 25 em bie't chai da cau, 30 em biet choi cau long , 15 em biet choi

ca hai Hoi lap lOAi c6 bao nhieu em chi bie't da cau? bao nhieu em chi

bie't danh cau long? ST so lop la bao nhieu?

Lai gidi

Dua vao bieu do ven ta suy ra so hoc

sinh chi bie't da cau la 25 -15 = 10

So hoc sinh chi bie't danh cau long la

30-15 = 15

Do do ta CO sT so'hoc sinh cua lap lOAi la

10 + 15 + 15 = 40

V i du 2: Trong lop IOC c6 45 hoc sinh trong do c6 25 em thich mon Van, 20

em thich mon Toan, 18 em thich mon Su, 6 em khong thich mon nao, 5

em thich ca ba mon Hoi so em thich chi mot mon trong ba mon tren

Loi gidi

Goi a,b,c theo thu t u la so hoc sinh chi thich mon Van, Su, Toan;

X la so'hoc sjnh chi thich hai mon la van va toan

y la so hoc sjnh chi thich hai mon la Su va toan

z la so hoc sjnh chi thich hai mon la van va Sir

Ta CO so em thich it nhat mot mon la 45 - 6 = 39

Sua vao bieu do ven ta c6 he phuong trinh

Bai 1.29: Mpt nhom hoc sinh gioi cac bp mon: Anh, Toan, Van Co 8 em gioi

Van, 10 em gioi Anh, 12 em gioi Toan, 3 em gioi Van va Toan, 4 em gioi Toan va Anh, 5 em gioi Van va Anh , 2 em gioi ca ba mon Hoi nhom do c6 bao nhieu em?

Huong dan gidi

Ky hi?u A la tap hop nhiing hoc sinh gioi Anh, T la tap hop nhCrng hoc sinh gioi toan, V la tap hop nhijng hoc sinh gioi Van

Bai 1.30: Co 40 hoc sinh gioi, moi em gioi it nha't mot mon Co 22 em gioi Van,

25 em gioi Toan, 20 em gioi Anh Co 8 em gioi diing hai mon Van, Toan; Co

7 em gioi dung hai mon Toan, Anh; Co 6 em gioi diing hai mon Anh, Van

Hoi: Co bao nhieu em gioi ca ba mon Van, Toan, Anh? (^f^ , i

Huang dan gidi *t ,

Ky hieu A la tap hop nhijng hoc sinh gioi Anh, T la tap hop nhCrng hoc sinh gioi toan, V la tap hop nhirng hoc sinh gioi Van

Theo gia thie't t a c 6 : n ( V ) = 22, n(T) = 25, n ( A ) = 20,

Bai 1.31: Trong Ky thi tot nghiep pho thong, o mot trirong i<e't qua so thi sinh

dat danh hieu xuat sac nhir sau; Vo mon Toan: 48 thi sinh; Ve mon Vat ly: 37

thi sinh; Ve mon Van: 42 thi sinh; Ve mon Toan hoac mon Vat ly: 75 thi sinh;

Ve m o n Toan hoac m o n Van: 76 thi sinh; Ve mon Vat ly hoSc m o n Van: 6(i

thi sinh; Ve ca 3 mon: 4 thi sinh Vay co bao nhieu hoc sinh nhan duoc danh

hieu xuat sac ve:

a) M o t mon?

b) Hai mon?

c) it nhat mot mon? if

•s**' Huong dan gidi

Ggi A, B, C Ian lugt la tap hop nhi>ng hoc sinh xuat sac ve mon Toan, mon

Vat Ly, mon Van

Goi a, b, c Ian lugt la so hoc sinh chi dat danh hieu xuat sac mot mon ve

mon Toan, mon Vat Ly, mon Van

Goi X, y, z Ian lugt la so hoc sinh dat danh hieu xuat sac hai mon ve mon

Toan va mon Vat Ly, mon Vat Ly va mon Van, mon Van va mon Toan

D i i n g bieu do Ven dua ve he 6 phuong trinh 6 an sau:

DS: a) 65 thi sinh dat danh hieu xuat sac 1 mon

b) 25 thi sinh dat danh hieu xuat sac 2 mon

c) 94 thi sinh dat danh hieu xuat sac it nha't 1 mon

D A N G T O A N 3: CHUNG MINH TAP HOP BANG NHAU, TAP HOP CON

V i d u 2: Cho A va B la hai tap hop Chung minh rang

c) Ta CO X e A >>J (B \) <=> X e A

x e ( B \ A )

x e A xeB » X e A

X G A ' X G B

a) Chung minh rang A = B b) A c C

Huang dan gidi

V o i x G A v i A c B = > x G B = > x G B u D Suy r a A u C c B u D

X G A b) Ta CO Vx, X G A n C <=> X G C X G A

Vi A c B => X G B Suy ra A n C c B

X G B c) VX, X G CgA u A <=> x G C B A

X G A <=> X ^ A O X G B

X G A Suy ra C B A U A = B

Bai 1.34: Cho cac tap hop A, B va C Chung minh rang

a) ( A \ B ) u ( B \ A ) = ( A u B ) \ ( A n B ) b) A \ ( B n C ) = ( A \ B ) u ( A \ C ) c) A \ ( B u C ) = ( A \ B ) n ( A \ C )

Huang dan gidi

fx G A ' X G A \ ^

l^ltan loai vd phumig phiipgiiii Uai so W

- Phan khong bj gach bo chinh la giao ciia hai tap h^p A, B

• Detim A^^B taiamnhusau ,^ t

- Sap xep theo thu tu tang dan cac diem dau mut cua cac tap hgpA, B len

true so ' ^ ^

- To dam cac tap A, B tren true so

- Phan to dam chinh la hop cua hai tap hop A, B ^

L f y INHH MI V UVVH KHang Vtei

• De tim A \ ta lam nhu sau

- Sap xep theo thu tu tang dan cac diem dau miit cua cac tap hgp A, B len true so ,^

- Bieu dien tap A tren true so (gach bo phan khong thuoe tap A ) , gaeh be phan thuoe tap B tren true so

- Phan khong bi gaeh bo chinh la A \ i r nt -r- •

ea 1 cAc VI D u MINH HOA

Vi du 1: Cho cac tap hop:

A = { x 6 R l x < 3 } B = { X G R I 1 < X < 5 } C = {X G R I - 2 < X < 4}

a) Hay viet lai cac tap hgp A, B, C duai ki hieu khoang, nira khoang, doan

b) Tim A u B, A n B, A \ c) Tim ( B u C ) \ ( A n C )

Vay m > - ^ la gia trj can t i m

ca 2 BAI TAP LUYEN TAP

Bai 1.35: Xac d j n h cac tap hgp A u B , A \ C , A n B n C v a bieu dien tren tryc so

cac tap hgp t i m dugc biet:

Bai 1.36: Cho hai tap hgp A = [0; 4), B = {x e IR /|x| < 2}

H a y xac d j n h cac tap hgp A u B , A n B , A \

Huang dan gidi

a) De (1; m ] n ( 2 ; +oo)^0 thi m < 2 b) Viet tap A g o m cac phan tix x thoa man dieu ki^n

35

u r y ii\nttMl V UVVH Khang vtft

b) A n B = 0 «

- 2 < m - 1 ,,

m + 1 ^ » - l < m < 3

< 2 m < 3 Ket hop voi dieu k i ^ n (*) ta c6 - 1 < m < 3 la gia t r i can t i m

Bai 1.40: Cho hai tap khac rong : A = ( m - 1 ; 4 ] , B = (-2 ; 2 m + 2 ) , v o i m e R Xac

d i n h m d e : j " : , '

a) A n B ^ 0 ; • b) A c B ; 1*

B e A; , , d) ( A n B ) c ( - l ; 3 )

' Huang dan gidi

\6i A = ( m - 1 ; 4 ] , B = (-2 ; 2 m + 2) khac tap rdng, ta c6 dieu k i ^ n

a) A n B ^ 0 « m - 1 < 2 m + 2 « m > - 3 So sanh v o i (*) ta thay cac gia trj m

thoa m a n yeu cau AnB^0 la - 2 < m < 5

UN A D f m - l > - 2 f m > - l

b) A e B c ^ l ^ ^ ^ ^ ^ ^ ^ l ^ ^ ^ < ^ m > l So s a n h n t a thay cac gia t r i m

thoa m a n yeu cau A e B la 1 < m < 5

Trong nhieu t r u o n g hop ta khong the biet dugc gia trj d u n g cua d^i l u g n g

ma ta chi bie't so gan d i i n g ciia no ,

^ V i d u : gia trj gan d i i n g ciia n la 3,14 hay 3,14159; con doi vai 72 la 1,41 hay

1,414;

N h u vay c6 s u sai l?ch giua gia tri chinh xac cua m o t dai l u g n g va gia t r i

gan d u n g ciia no De danh gia miic dg sai l#ch do, n g u o i ta dua ra khai

niem sai so tuyet do'i

Sai so tuyet doi:

Sai so t u y ^ t d o i ciia so gan d u n g Neu a la so gan d i i n g ciia a thi ^^ = so'gan d i i n g a , (,,( »*•

D p c h i n h xac cua m p t so gan d i i n g

Trong thuc te^ nhieu k h i ta khong bie't a nen ta khong tinh dugc Ag nhien ta c6 the danh gia Ag khong v u g t qua mot so d u a n g d nao do

Neu Aa < d thi a - d < a < a + d , k h i do ta viet a = a ± d

Goi la do chinh xac cua so gdn dung j

Sai so tucmg d o i

Sai sotuang do'i ciia so gan d i i n g a, k i hieu la 63 la ti s o g i i i a sai so'tuyet

dugc ggi la sai so'tuyet dot ciia

chat l u g n g ciia phep do dac hay tinh toan cdng cao

Quy tron so gdn dung

N g u y e n tac q u y t r o n cac so n h u sau:

Neu c h i i so ngay sau h a n g q u y tron nho han 5 thi ta chi viec thay chir so' do

va cac chir so'ben phai no boi 0

Neu chu so ngay sau hang quy tron Ion hon hay b i n g 5 thi ta thay chij so do

va cac c h i i soben phai no boi 0 va cong them mot don vj vao sohang lam tron

Nhdn xet: K h i thay so d i i n g boi so q u i tron den mot hang nao do t h i sai so'

tuyet doi ciia so' q u i tron khong v u g t qua nua d o n v i ciia hang q u i tron

N h u vay, do chinh xac ciia so q u i tron bang nua don vj ciia hang q u i tron

Chii y: Cac viet so quy tron oi a so gan dung can c i i vao dp chinh xac cho truoc:

Cho so' gan d i i n g a v o i do chinh xac d K h i dugc yeu cau quy tron a ma khong noi ro quy tron den hang nao thi ta quy tron a den hang cao nha't ma

d n h o hem m p t d a n v i ciia hang do

Chu sochdc (dang tin)

Cho so gan d i i n g a ciia so a v a i do chinh xac d Trong so' a, mgt c h i i so' dugc ggi la c h u so c h i c (hay dang tin) neu d khong v u g t qua nua d a n vj ciia hang c6 chir so' do : „

Nhdn xet: Tat ca cac chCr so d i i n g ben trai chi i so'chac deu la chi i so chac Tat

ca cac chu so d i i n g ben phai chii so khong chac deu la c h i i so khong chSc

©ang chudn cua so gdn diing

Neu so gan d i i n g la sothap phan khong nguyen thi dang chuan la dang ma

m g i c h i i so'ciia no deu la c h i i chac chan / j : ,

- Neu so'gan d i i n g la so nguyen thi dang chuan ciia no la A.10*^ trong do A

la so nguyen, k la hang thap nha't c6 chu sochSc ( k e N ) (suy ra moi chii so

cua A deu la c h u so'chic chan) ,y

Khi do d o chinh xac d = CS.IO""

6 JCi hieu khoa hoc cua mot so

MQ\O thap phan khac 0 deu vie't dugc dudi dang a.lO", 1 < |a| < 10, n e Z

(Quy uoc 10"" = ) dang n h u vay dugc goi la kihieu khoa HQC cua so do

B CAC DANG TOAN VA PHlTaNG PHAP G I A I

DANG TOAN 1: TINH SAI SO TUYETDOI, SAI SO TUONG DOI CUA SO GANDUNG VIET SO QUY TRON

ea 1 cAc v f D u MINH HOA

V i 1: D o dai cua cai cau ben thuy hai (Nghe A n ) nguai ta d o dugc la

996m ± 0,5m • Sai so tuang doi to! da trong phep do la bao nhieu

Loi gidi

Ta CO do dai gan d u n g cua cau la a = 996 vol do chinh xac d = 0,5

V i sai so tuy^t doi Ag < d = 0,5 nen sai so tuong doi

|a| a| 996 Vay sai so tuang doi toi da trong phep do tren la 0,05%

V i dv 2: Hay xac d i n h sai so tuyet doi cua cac so gan d u n g a, b biet sai so

tuong doi cua chung

a) Ta CO 0,001 < 0,002 < 0,01 nen hang cao nhat ma d nho hon mot don vj cua

hang do la hang phan tram

Do do ta phai quy tron so a = 2,235 den hang phan tram suy ra a * 2,24 b) Ta CO 100 < 101 < 1000 nen hang cao nhat ma d nho h o n m o t d o n v i ciia hang do la hang nghin :» it • , - :

Do do ta phai quy tron so a = 23748023 den hang nghin suy ra a a; 23748000

V i du 4: a) Hay vie't gia trj gan diing ciia \/8 chinh xac den hang phan tram

va hang phan nghin biet Ts = 2,8284 l/oc lugng sai so tuy^t doi trong

>:\i

Do do gia trj gan diing ciia N/2015'' den hang chuc la 25450

= 72015'' - 25450 < 25450,72 - 25450 = 0,72

Ta CO 72015^ -25450

Suy ra sai so tuyet doi ciia so gan diing 25450 khong v u g t qua 0,72

Gia trj gan d i i n g ciia 72015'* den hang tram la 25500

Ta CO 72015" - 25500 = 25500 - ^2015" < 25500 - 25450,71 = 49,29

Suy ra sai so tuy^t doi ciia so gan diing 25500 khong v u g t qua 49,29

ea 2 BAI T A P LUVeN T A P Bai 1.41 Su d u n g may tinh bo tiii, hay vie't gia trj gan diing ciia m o i so sau,

chinh xac den hang phan tram va hang phan nghin :

Huong dan gidi

a) Sit dung may tinh bo tiii ta c6 N/S = 1,732050808 Do do: Gia tn gan diing

ciia V3 chinh xac den hang phan tram la 1,73 Gia tri gan dung ciia -JS

chi'nh xac den hang phan nghin la 1,732

b) Sit dung may tinh bo tiii ta c6 gia tn cua la 9,8696044 Do do : Gia trj

gan diing ciia chinh xac den hang phan tram la 9,87 Gia tri gan diing

cua chinh xac den hang phan nghin la 9,870

Bai 1.42: Hay vie't so'quy tron cua so a voi do chinh xac d dugc cho sau day;

a) a = 17658 ± 16 ; b) a =15,318 ±0,056

Huang dan gidi

a) Vi 10 < 16 < 100 nen hang cao nha't ma d nho han mot don vj cua hang do la

hang tram Nen ta phai quy tron so 17638 den hang tram Vay so quy tron la

17700 (hay vie't a * 17700)

b) Ta CO 0,01 < 0,056 < 0,1 nen hang cao nha't ma d nho hon mot don vj ciia

hang do la hang phan chuc Do do phai quy tron so 15,318 den hang phan

chuc Vay so quy tron la 15,3 (hay vie't a «15,3)

2 Bai 1.43: Cho so x = - Cho cac gia trj gan diing ciia x la :0,28 ; 0,29 ; 0,286

Hay xac dmh sai so tuyet doi trong tung truong hop va cho biet gia trj gan

diing nao la tot nha't

Huang dan gidi

Ta CO cac sai so tuyet doi la :

0,28

7 175 A u = 0,29 7 700 0,286 3500

Vi Ac < Ab < An nen c = 0,286 la so gan diing tot nha't

Bai 1.44: Mot mie'ng da't hinh chii nhat c6 chieu rpng x = 43m ± 0,5m va chieu

daiy = 63m ± 0,5m

Chung minh ring chu vi P ciia miehg da't la P = 212m ± 2m

Huang dan gidi

Gia sir X = 43 + u, y = 63 + V

Taco P = 2x + 2y = 2(43 + 63) + 2u+2v = 212 + 2(u + v )

Theo gia thie't - 0 , 5 < u < 0 , 5 v a -0,5<v<0,5 nen -2 < 2(u + v) < 2

Dodo P = 2 1 2 m ± 2 m

DANG TOAN 2: XAC DINH CAC CHU SO CHAC CUA MOT

SO GAN DUNG, DANG CHU AN CUA CHU SO GAN DUNG VA Ki HIEU KHOA HOC CUA MOT SO

con chii so hang nghin(so 4) la chii so chac

Vay chu so chac la 1,2,3,4 Cach vie't duoi dang chuan la 3214.10'^ b) Ta C O 6a = ^ Aa =5a.|a| = l%.l,3462 = 0,013462

Suy ra do chinh xac ciia so gan diing a khong vugt qua 0,013462 nen ta c6 the xem do chinh xac la d = 0,013462

Ta CO = 0,005 < 0,013462 < ^ = 0,05 nen chu so hang phan tram(so 4)

khong la so chac, con chu so hang phan chiic(so 3) la chii so chac i) ( Vay chij' so chac la 1 va 3

Cach vie't duoi dang chuan la 1,3

Vi d\i 2: Viet cac so gan diing sau duoi dang chuan

p._£huso chir so chac do do so gan diing vie't duoi dang chuan la 2,5

du 3: Cac nha khoa hoc My dang nghien ciiu lieu mot may bay c6 the c6 toe do gap bay Ian toe do anh sang Voi may bay do trong mot nam (gia

su mot nam c6 365 ngay) no bay dugc bao nhieu? Biet van toe anh sang la

- ^jgo nghin km/s Viet ke't qua duoi dang ki hieu khoa hoc ^

Lai gidi

Ta C O mot nam c6 365 ngay, mot ngay c6 24 gio, mpt gio c6 60 phut va mot

phiit C O 60 giay

Vay mot nam c6 24.365.60.60 = 31536000 giay

V i van toe anh sang la 300 nghin km/s nen trong vong mot nam no di dupe

31536000.300 = 9,4608.10'^ k m ; ; -i;-:::;

ca 2 B A I T A P L U Y E N T A P

Bai 1.45: So dan cua mot tinh la A = 1034258 ± 300 (nguoi) Hay tim cac chi> so

chac va viet A d u o i dang chuan

Hu&ng dan gidi

Ta C O : = 50 <300 < 500 = 1^9^ nen cac chu so 8 (hang don vj), 5 (hang

chuc) va 2 ( hang tram ) deu la cac chir so khong chac

Cac chCr so con lai 1, 0, 3, 4 la chi> so chac

Do do each viet chuan cua so A la A « 1034.lO'' (nguoi)

Bai 1.46: Do chieu dai cua mot con doc, ta dupe so do a = 192,55 m , voi sai so

tuang doi khong vupt qua 0,2% Hay tim cac chu so chac cua d va neu each

viet chuan gia trj gan dung cua a

Huang dan gidi

Ta C O sai so t u y f t doi cua so do chieu dai con doc la :

Aa =a.5-, < 192,55.0,2% = 0,3851

Vi 0,05 < Aa < 0,5 Do do chu so chac cua d la 1, 9, 2

Vay each viet chuan cua a la 193 m (quy tron den hang don vi)

Bai 1.47: Cho 3,141592 < 7t < 3,141593 Hay viet gia trj gan diing cua so TI

duoi dang chuan va danh gia sai so tuy?t doi ciia gia trj gan dung nay tron^

moi truong hop sau :

a) Gia tri gan dung cua TI c6 5 chii' so chac;

b) Gia trj gan d u n g cua n c6 6 chir so chac;

c) Gia trj gan diing cua 7t c6 3 ehi> so'chac

Huang dan gidi

a) V i C O 5 chu- so chac nen so gan diing cua n dupe viet d u o i dang chuan la

3,1416 (hay T I * 3 , 1 4 1 6 )

Sai s o t u y e t d o i c i i a so gan dung la = |3,1416 - TI| < 0,000008

b) V i C O 6 chir so chac nen TI a 3,14159 va sai so tuyet doi cua so'gan dung na\

la A„ = |3,14159 - Ti| < 0,000003

c) V i c 6 3 e h i i s o c h a c n e n T i * 3 , 1 4 va A^ |3,14 - n| < 0,001593

O N T A P C H U O N G I Bai 1-48: Cho O x y , lap menh de keo theo va nienh de tuong d u o n g ciia hai

menh de sau day va cho biet tinh dung, sai ciia chiing:

p : "Diem M nSm tren phan giac cua goc Oxy "

n • "Diem M each deu hai canh Ox, O y " "' " ""'' '

Bai 1.49: Cho djnh li : "Cho so t u nhien n Neu n"^ chia he't cho 5 thi n chia het

cho 5" Dinh li nav dupe viet duoi dang P => Q

a) Hay xac dinh cac menh de P va Q

b) Phat bieu dinh li tren bang each diing thuat ngir "dieu kien can"

c) Phat bieu dinh li tren bang each diing thuat ngu "dieu kien d i i " d) Hay phat bieu djnh li dao (neu c6) cua dinh li tren roi diing cac thuat ngij

"dieu ki^n can va d i i " phat bieu gpp ca hai dinh li thuan va dao

Huong dan gidi

a) P : " n la so t u nhien va n'^ chia he't cho 5", Q : " n chia he't cho 5"

b) Voi n la so t u nhien, n chia he't cho 5 la dieu kien can de n'^ chia he't cho 5 ; hoac phat bieu each khae : Voi n la so t u nhien, dieu kien can de chia he't cho 5 la n chia he't cho 5

c) Voi n la s o t u nhien, n"* chia he't cho 5 la dieu kien du de n chia he't cho 5

Djnh li dao : "Cho so'tu nhien n, neu n chia he't cho 5 thi n^ chia he't cho 5"

That vay, neu n = 5k thi n"^ = 5\k'^: So nay chia he't cho 5

Dieu kien can va du de n chia he't cho 5 la n"* chia he't eho 5 ' '' •'

Bai 1.50: Cho tap X = { 1 ; 2 ; 3 ; 4 ; 5 ; 6 ; 7 }

3) Hay tim tat ca cac tap con ciia X c6 chiia cac phan tir 1, 3, 5, 7

°) Co bao nhieu tap con ciia X chira diing 2 phan tir ?

Huang dan gidi

^) Cac tap eon ciia X chira c6 cac phan tir 1, 3, 5, 7 dupe thanh lap bang each them vao tap { 1 ; 3 ; 5 ; 7 } cac phan tir con lai ciia tap X '

Do do tat ca cac tap con cua X c6 chiia cac phan t u 1, 3, 5, 7 la :

{ 1 ; 3 ; 5 ; 7 } , { l ; 3 ; 5 ; 7 ; 2 } , { ] ; 3 ; 5 ; 7 ; 4 } , { l ; 3 ; 5 ; 7 ; 6 } ,

{ 1 ; 3 ; 5 ; 7 ; 2 ; 4 } , { 1 ; 3 ; 5 ; 7 ; 2 ; 6 } , { 1 ; 3 ; 5 ; 7 ; 4 ; 6 } va X

b) Gia s u tap can t i m la i a ; b j v o i a b ., - -rv''•

• V i X CO 7 phan t u nen c6 7 each chon phan tir a Sau k h i chon a t h i X con 6

phan t u , do do v o i moi each chon a, ta c6 6 each chon phan t u b, n h u vay c6

7.6 = 42 cap (a ; b) theo each chon nay

N h u n g v o i each chon tren t h i v o i hai phan t u bat k"i a, b ta da chon lap lai

hai Ian, do la hai cap {a;b) va (b;a), n h u n g chi c6 d u y nhat t a p { a ; b }

42

Do do, CO —="1^ tap con cua X chua d u n g hai phan t u

Bai 1.51: Xet t i n h d i i n g sai ciia menh de sau va neu m?nh de p h u d j n h ciia no

a) 3 x € Q : 4 x ^ - 1 = 0 ; b ) 3 x G Z , x ^ = 3 ;

c) V n G N * : 2 " + 3 la m o t so nguyen to ; d) Vx e M, x^ + 4x + 5 > 0

e) V X G I R , X ' * - x ^ + 2 x + 2 > 0

' Huang dan gidi

a) Giai p h u o n g t r i n h : 4x^ - 1 = 0 o x = ± i e Q Vay menh de da cho d u n g

M e n h de p h i i d j n h \/x&Q: Ax^

-b) Ta CO x^ = 3 <=> x = ±V3 V i ±N/3 i Z nen menh de da cho sai

M e n h de p h i i d j n h Vx G Z, x^ ;^ 3

c) V o i n = 5 t h i 2 " + 3 = 35, so nay chia het cho 5 (khong nguyen to) D o do

menh de da cho sai

M e n h de phii d j n h " B n e N * : 2" + 3 khong phai la m o t so nguyen to"

b) D j n h l i tren c6 d j n h If dao k h o n g ? Giai thich

c) Su dung thuat ngi> "dieu ki^n can" va "dieu kien d i i " de phat bieu djnh If tren

Huang dan gidi

a) x e M , x < 0 ^ x + l < - 2 ^ ' b) M e n h de dao la x G K, x + - < - 2 => x < 0 ".^ •

C h u n g m i n h rang v o i n G N, ta co: n : le o 3n + 1 : chan

Huang dan gidi

Thmn: Cho n : le, thi n = 2k + l {k eN) V , ! ,V, ,

r:> 3« + 1 = 3(2^ + l) + -\=6k + 4 = 2{3k + 2), v o i 3^ + 2 G N

=> 3« + 1: chan

Dao : Cho 3n + 1 : chan, ta chung m i n h n : le

D u n g p h u o n g phap phan chung: )

Gia s u n : chan, tuc \an = 2k(k e N) * j

=> 3« + 1 = 6A: + 1 => 3?7 + 1 : le: trai v o i gia thie't Vay 3n + l: le '^^

T u hai phan thuan va dao ta duoe: « : le o 3« + 1: chan Bai 1.54: Cho cac tap hop:

A = { x G Z i l < x < 6 ) ,

B = { X G Q I ( 1 - 3 X ) ( X ' ' - 3 x ^ + 2 ) = 0|, C = (0;l;2;3;4;5;6} *

a) Viet cac tap h o p A, B d u o i dang liet ke cac phan t u , tap C d u o i dang chi ro

tfnh dac t r u n g cua phan t u b) T i m A n B , A u B , A \ B , C g ^ ^ A A n B c) C h u n g m i n h rang A n ( B u C ) = A

* Huang dan gidi

fiian loai va pnwimg pnapgiiu tjiii so lu

Bai 1.55: T i m quan he bao ham hay bang nhau giua cac tap hop sau day:

b) Hay xac d j n h tat ca cac tap P biet rang (A n B) c P c (A u B)

Huang dan gidi

Bai 1.57: Cho ba tap hop:

A = { x e R | - 3 < x < l } ; B = { x 6 R | - l < x < 5 | ; C = {x e S ||x| > 2' a) Xac djnh cac tap hop sau day va viet ket qua d u o i dang khoang, doan hay nua khoang: A o B, A u B, (B \) n C

(1) v a ( 2 ) c h o : C^ ( A u B) = ( C B { A ) n ( C K B ) Bai 1.58: Cho hai tap hop A, B bat k i

C h u n g m i n h rang: A u B = A r i B < = > A = B

Huattg dan gidi

• Thuan:/4 u B = /4 n B, ta chung m i n h :/4 = B

Vx, X e A X € A u B (vi A c A u B) => X e A n B ( v i A u B = A o B ) x € B (vi A n B c B)

N h u t h e : Vx,x e A => X e B, nen A c B (a) , >

V x , x e B= > x G A u B ( v i B c A u B ) = > x € A n B ( v i A u B = A n B ) => x e A (vi A n B c A)

N h u t h e : V x , x e B = > x e A , n e n B c A (b) i> Tir (a) va (b) cho/I = B

• Dao: Cho/4 = B, ta Chung m i n h : A u B = A n B '

%hu(mq2 H A M S O B A C N H A T V A B A C H A I

§l.DAICUaNGVEHAMS6 m - ^ -.^.i

A T O M T A T L Y T H U Y E T * ^

L ©mn ngnia

• Cho D c K , D 7i 0 Ham so i xac djnh tren D la mot qui t3c dat tuong un,

moi so' X 6 D voi mot va chi mot so y e

• X dugc goi la bien so (doi so), y dugc gpi la gia tri cua ham so / tai x

Ki hieu: v = f (x)

• D dugc goi la tap xdc djnh ciia ham so f K

2 Cdch cho ham so

• Cho bang bang • Cho bang bieu do • Cho bang cong thiic y = f(x)

Tap xdc dinh cua ham so y = f (x) la tap hop tat ca cac so thuc x sao chi

bieu thuc f(x) co nghla

Do thi ciia ham so y = f (x) xac dinh tren tap D la tap hop tat ca cac dicn

M(x; f(x)) tren mat phang toa do voi mgi x e D

Chu y: Ta thuong gap do thi cua ham so' y = f (x) la mot duong Khi do t

noi y = f (x) la phuattg trinh cua duong do

4 Su bien thien cua ham so ^

Cho ham so f xac dinh tren K ' ^

• Ham so y = f (x) dong bien (tang) tren K neu

Vxi,X2 e K : Xj < X2 => f(xi) < f ( x 2 )

• H^m so y = f(x) nghich bien (giam) tren K neu

v-Vxi,X2 G K : X I < X 2 =>f(x^)>f(x2) ^

5 Tinh chdn le cua ham so

Cho ham so y = f (x) c6 tap xac dinh D

• Ham so f dugc goi la ham so chin neu voi Vx e D thi -x € D va f(-x) =f(x!

• Ham so f dugc ggi la ham so le neu voi Vx e D thi -x G D va f(-x) =:-f(x)

•^U Chii y: + Do thi cua ham sochan nhqn true tung lam true doi xi'mg

+ Do thi eua ham sole nhqn goe toq do lam tam doi xung

6: Tinh tien do thi song song vcri true toa do ^

Dinh ly: Cho (G) la do thj cua y = f (x) va p > 0, q > 0; ta c6

Tjnh tien (G) len tren q don vj thi dugc do thi y = f (x) + q Tinh tien (G) xuong duoi q don vj thi dugc do thi y = f (x) - q Tinh tien (G) sang trai p don vi thi dugc do thi y = f (x + p) Tjnh tien (G) sang phai p don vi thi dugc do thj y = f (x - p)

7P(X) CO nghla o P(x) > 0

CO nghla <=> P(X) >0

ca 1 CAC Vf DU MINH HQA

Vi du 1: Tim tap xac dinh cua cac ham so sau

Suy ra tap xac dinh cua ham so la

j 5 ^ j ^ J V 2 - V 6 V 2 + V 6 -V2->/6 -V2 + 76

Vi d\ 2: Tim tap xac dinh cua cac ham so sau

a) y = V5-3|x|

X + 4

4^- 16 Lot gidi

Suy ra tap xac dinh cua ham so la D = (-oo; -4) u (4; +oo)

Vi d\ 3: Tim tap xac djnh ciia cac ham so'sau

a) DKXD: x2+2x + 3^0 diing voi mgi x

Suy ra tap xac djnh ciia ham so la D = M

b) Khi X > 1 thi ham so la y = - luon xac dinh voi x > 1

Do do ham so da cho xac djnh khi x > -1

Suy ra tap xac dinh cua ham so'la D = [-l;+oo)

Vi d\ 4: Cho ham so: y - • mx

Vx - m + 2 - 1 a) Tim tap xac djnh ciia ham so theo tham so m

b) Tim m de ham so xac dinh tren (0;1)

voi m la tham so'

Vay m G (-OO;1] U {2} la gia trj can tim

p 2 BAI TAP LUYIN TAP

Bai 2.0 Tim tap xac dinh ciia cac ham so sau:

b) y = Vx + 2 d) y = X + Vx^ -4x + 4

X - 3Vx + 2

f x 2 - 2 x + 3>0 a) DKXD: \

Bai 2.2: Tim gia tri ciia tham so'm de:

\- x + 2 m + 2 , ^ , - / , „ s

a) Ham so y = xac dinh tren (-1;0)

b) H a m so y =

X - m

CO tap xac djnh la [0;+oo)

Huang dan gidi

Neu m > 0 thi n « X > m D = [m;+oo) nen m > 0 khong thoa man

Neu m < 0 thi (*) o x > 0 D = [0; +oo)

Vay m < 0 la gia trj can tim

Bai 2.3: T i m gia trj cua tham so m de:

a) H a m so y = > y x - m + l + 2x

V-x + 2m xac dinh tren ( - 1 ; 3 ) b) H a m so y = Vx + m + V 2 x - m + l xac dinh tren (0; +oo)

^ xac djnh tren ( - 1 ; 0 ) c) H a m so y = V-x - 2 m + 6 -

S u dung dinh nghia

H a m so y = f(x) xac dinh tren D :

[Vx e D =i> - x e D

f ( - x ) = f(x)

Vx e D => - x e D f(-x) = - f ( x ) •

Chii y : M p t ham so c6 the khong c h i n cung khong le

Do thi ham so chan nhan true Oy lam true doi xung

Do thi ham so le nhan goc tpa do O lam tam doi xiing

Q u y trinh xet ham so chan, le

31: T i m tap xac djnh ciia ham so

r t y TNHHMTVDWH KRangYiet

B2: Kiem tra •

Neu Vx e D = > - X e D Chuyen qua buoc ba

Neu 3X(, e D => -x,, i D ket luan ham khong c h i n cung khong le ^

B 3 : x a c d i n h f ( - x ) v a s o s a n h v a i f ( x ) Neu bang nhau thi ket luan ham so la chan ,

Neu doi nhau thi ket luan ham so la le

Neu ton tai mot gia tri BXQ e D ma f ( - X Q ) f ( X Q ) , f ( - X Q ) ^ - f ( X Q ) ket luan

ham so khong c h i n cung khong le

D o d o f(x) = 3 x ^ + 2 ^ l a h a m s o l e 4 b) Ta C O TXD: D = K

Voi moi xelR ta c6 - x e M va f(-x) = {-xf + ^{-xf +1 = x"* + Vx^ +1 = f(x)

Phdtt loai vd phuong phdp gidi Dai so 10

p 2 BAI TAP LUYEN TAP

Bai 2.4:: Xet tinh chin, le cua cac ham so sau:

a) f(x) = > / ^ - V T ^ b ) f ( x ) = c) f(x) = 3x2-2x4-1 d ) f ( x ) =

x - 5

x - 1 x^

-Vay ham so da cho la ham so le

Bai 2.5: Cho ham so y = f (x), y = g(x) c6 cung tap xac dinh D Chiing minh rang

a) Neu hai ham so tren le thi ham so y = f(x) + g(x) la ham so'le

b) Neu hai ham so'tren mpt chan mot le thi ham so y = f (x)g(x) la ham so le

Huang dan gidi

a) Ta CO ham so y = f (x) + g (x) c6 tap xac djnh D

Do ham so' y = f (x), y = g(x) le nen Vx e D => -x e D

DANG TOAN 3: XET TINH DONG BIEN, NGHICH BIEN(DON

Dim cm HAM so TREN MOT KHOANG

(^huang phdp gidi

CI: Cho ham so y = f(x) xac djnh tren K

Lay x , , X 2 eK; x, < X2 , dat T = ((xj) - f(xi)

• Ham so dong bien tren K <=> T > 0

• Ham so nghich bie'n tren K o T < 0 C2: Cho ham so y = f(x) xac djnh tren K

^ f ( x 2 ) - f ( x i ) Lay X p X 2 eK; x^ ^ X 2 , dat T = — ^

X2 X]

• Ham so dong bie'n tren K <:> T > 0

• Ham so nghjch bie'n tren K <=> T < 0

m 1 CAC Vf DU MINH HOA

Vi du 1: Xet su bie'n thien cua ham so sau tren khoang (!;+«)

Phan loai va phttcmg phdp giai Dai so 10

a) Xet chieu bien thien cua ham so tren (-00;0) va tren (0;+oo)

b) Lap bang bien thien cua ham so tren [-1;3] tir do xac d i n h gia t n Ion

nhat, nho nhat cua ham so tren [-1;3]

Lm Giai

T X D : D = R

a) V x i , X 2 € 1R,X] < X 2 X 2 - x , > 0

Ta c 6 T = f ( x 2 ) - f ( x , ) = ( x ^ - 4 ) - ( x ? - 4 ) = x 2 - x 2 = ( x 2 - X i ) ( x i + X 2 )

Neu X i , X 2 e (-00;0) T < 0 Vay ham so y = f (x) nghich bien tren (-00;0)

Ne'u X i , X 2 e (0;+oc) T > 0 Vay ham so y = f (x) dong bien tren (0;+oo)

b) Bang bien thien cua ham so y = x^ - 4 tren [-1;3]

y = x ^ - 4 -3

-4 - ^ " ^

5

Dua vao bang bien thien ta c6

m a x y = 5 k h i va chi khi x = 3, m i n y = -4 khi va chi k h i x = 0

Suy ra p h u o n g trinh 74x + 5 + N/X - 1 = 3 v6 nghiem Voi X = 1 de tha'y no la nghiem cua phuong trinh da cho

Vay p h u a n g trinh c6 nghiem duy nhat x = 1

b) D K X D : X > 1

Dat x^ + 1 = t, t > 1 x^ = t - 1 p h u a n g trinh tra thanh

7 4 7+ 5 + 7 x ^ = 74t + 5 + 7 t ^ « f ( x ) = f ( t ) Ne'u x > t = > f ( x ) > f ( t ) hay 74x + 5 + 7x - 1 > 74t + 5 + 7 t ^ Suy ra p h u a n g t r i n h da cho v6 nghiem

Neu x < t = > f ( x ) < f ( t ) hay 74x + 5 + 7 x ^ < 7 4 t + 5 + 7 1 ^ Suy ra p h u a n g trinh da cho v6 nghiem

V^y f (x) = f ( t ) <=> X = t hay x^ + 1 = x o x^ - x + 1 = 0 (v6 nghiem)

Vay p h u a n g trinh da cho v6 nghiem

N h a n xet:

• H a m so y = f (x) dong bien (hoac nghich bien) thi p h u a n g trinh f (x) = 0 c6 tol da mot nghiem

• Neu ham so y = f(x) dong bien (nghjch bien) tren D thi f ( x ) > f ( y )

O x > y ( x < y ) va f(x) = f(y) x = y V x , y € D Tinh cha't nay duoc su

d u n g nhieu trong cac bai toan dai so n h u giai p h u a n g trinh , bat p h u a n g trinh , h^ p h u a n g trinh va cac bai toan cue trj

an loai va phucrng phdp giiii Dai so

ea 2 B A I T A P L U Y E N T A P

Bai 2.7: Xet su bie'n thien ciia cac ham so sau:

a) y = + 4x - 5 tren (-cc; -2) va tren (-2; +oc)

Xi,X2 6 (-so;-2) => K < 0 siiy ra ham so nghich bien tren ( - x ; - 2 )

X],X2 G (-2;+oc) r:> K > 0 suy ra ham so dong bien tren (-2;+oo)

b) Voi moi x,, X2 e S, x | X2 ta c6

2 2 _ 2 ( x , - X 2 ) 2

f ( x 2 ) - f ( x , ) =

X 2- 2 x , - 2 ( x 2 - 2 ) ( x , - 2 ) ( x 2 - 2 ) { x , - 2 ) Voi Xj,X2 G ( - x ; 2 ) => K < 0 do do ham so nghjch bien tren ( - x ; 2 )

Voi X ] , X2 G (2; +=o) => K < 0 do do ham so nghjch bien tren (2; +oo)

Vay ham so nghjc bien tren ( - = c ; - l ) - /

Bai 2.8: Chiing m i n h rang ham so y = x'' + x dong bien tren

Ap dung giai phuong trinh sau x"' - x = v/2x + 1 + 1

^ Humg dan gidi

Dat \/2x + 1 = y , phuong trinh tro thanh x"' + x = y^ + y

Qo ham so f (x) = x'' + x dong bien tren R nen

r X = -1

x = y=i>^2x + l = x < = > x - ^ - 2 x - l = 0 c = _ i + 75

2 Bai 2.9: Cho ham so y = % / x - l + x^ - 2 x

a) Xet su bien thien cua ham so da cho tren [l;+=o)

b) Tim gia trj ion nha't nho nha't ciia ham so tren doan [2; 5]

Huong dan gidi

a) Voi moi x ^ x j e[V,+x), Xj ^ Xj ta co

f ( x 2 ) - f ( x ! ) = ( N / ^ + x i - 2 x 2 ) - ( V ^ + x ? - 2 x , )

X2 - X,

7x2 - 1 + 7 x i - 1 + {X2 - X , ) ( X 2 + X i - 2 )

X2 - X , ^X2 - 1 +7X1 - 1 ^

Do do ham so da cho dong bien tren [!;+«:)

b) Ham so da cho dong bien tren [ l ; + x ) nen no dong bien tren [2;5]

Vay m a x y = y ( 5 ) = 17 o X = 5, m i n y = y(2) = 1 <=> x = 2

[2;S] [2;3]

D A N G T O A N 4: DO THI CUA HAM SO VA TINH TIEN DO THI HAM SO Phuong phdp gidi

• Cho ham so y = f(x) xac djnh tren D, Do thj ham so f la tap hop tat ca cac

diem M(x;f(x)) nam trong mat phang toa do voi x G D C/iMi/;Diem M(X(); y,,) e ( C ) _ do thj ham so y = f(x)<:=>yo = f ( x o )

• Su dung djnh ly ve tjnh tie'n do thi mot ham so'

Ql.CAC VI'DU MINH HOA

x^ +1 khi x > 2

du 1: Cho hai ham so f (x) 2x^ + 3x + 1 va g ( x ) = • 2 x - l khi - 2 < x < 2

6 - 5 x khi x < - 2 a) Tinh cac gia tri sau f ( - l ) va g ( - 3 ) , g ( 2 ) , g ( 3 )

^)Th-nx khi f ( x ) = l g ( x ) = l

Lbi gidi

^) Taco f ( - i ) = 2 ( - l ) 2 + 3 ( - 1 ) + ] = : 0 , g(-3) = 6 - 5 ( - 3 ) = 2 1 , g(2) = 2 2 - l = 3, g(3) = 3 2 + l = 1 0

At

a) T i m m d e d i e m M ( - l ; 2 ) thuoc do thi ham so'da cho

b) T i m cac diem co d j n h ma do thj ham so da cho luon d i qua v o i moi m

' Lcn gidi

a) D i e m M ( - l ; 2 ) thupc do thi ham so'da cho k h i va chi k h i

2 = - m - 2(m^ +1) + 2m^ - m <=> m = -2 Vay m = -2 la gia trj can t i m

b) De N ( x ; y ) la d i e m co'dinh ma do thj ham so da cho l u o n d i qua, dieu kien

can va d u la y = mx'^ - 2(m^ + l)x^ + 2m^ - m , V m

<=> 2m^ [ l - x^) + m (x^ - 1 j - 2x^ - y = 0, V m

1 - x ^ = 0

r x = i [ y = -2 2x^ + y = 0

Vay do thj ham so da cho luon d i qua diem N ( l ; - 2 )

Chii y: Ne'u da thuc a^x" +an_jx""^ + + aiX + ao = 0 v o i m o i x e K k h i va

chi k h i a„ =a„_i = = ao

Gpi M , N d o i x u n g nhau qua goc tpa dp O M ( x o ; y o ) => N ( - X o ; - y o )

V i M , N thupc do thj ham so nen y,, = - x ^ + x g + 3 x o - 4

- y o =x^ + x ^ - 3 x o - 4 _ f yo = -4 + 4+ 3 X ( ) - 4 j y o = -x^ + x^ + 3xo - 4

Vay hai diem can t i m c6 tpa dp la ( 2 ; - 2 ) va ( - 2 ; 2 )

ca 2 BAI TAP L U Y ^ N TAP

Bai 2.10: Cho ham so y = f (x) = -3x^ + m^x + m + 1 (voi m la tham so) a) T i m cac gia trj cua m de f (0) = 5

b) T i m cac gia trj cua m de do thj ciia ham soy = f ( x ) d i qua d i e m A ( 1 ; 0 )

Huang dan gidi

Bai 2.11: Tim cac diem co djnh ma do thj ham so sau luon di qua vai moi m

Vay diem can tim la A ( 2 ; 0 )

b) Diem co djnh la A (-4;-3), B(0;1)

Bai 2.12: a) Tjnh tien do thj ham so y = -x^ + 2 lien tiep sang trai 2 dan vi va

xuong duoi ^ don vj ta dugc do thj ciia ham so nao?

b) Neu each tinh tien do thi ham so y = x'' de dupe do thi ham so

y = x^+3x^+3x + 6

Huong dan giai

a) Ta tjnh tien do thi ham so y = -x^ + 2 sang trai 2 dan vj ta dupe do thj ham

so y = -(x + if + 2 roi tinh tien len tren - dan vi ta duac do thi ham so

2 • • •

y = - ( x + 2 ) 2 + 2 + l

b) Ta C O x^ + 3x^ + 3x + 6 = (x +1)^ + 5

Do do tinh tien do thj ham so y = x"' de dupe do thj ham so

y = x"' + 3x^ + 3x + 6 ta lam nhu sau

Tjnh tien lien tiep do thj ham so y = x'' di sang ben trai 1 don vj va len tren

y = ax + b ( a > 0 )

- 0 0

y = ax + b ( a < 0 )

- 0 0

3 ^ ( 5 thi

Do thi cua ham so y = ax + b {a*^ 0) la mot duong thang co h^ so goc bang

^ " va true tung tai B(0;b)

a, cat trvic hoanh tai A - - ; 0

• Cho duong thSng d co h^ so goc k, d di qua diem M(xo;yo)/ khi do

phuong trinh cua duong thang d la: y - yg = a(x - XQ )

B CAC DANG TOAN VA PHl/QNG PHAP G I A I

DANG TOAN 1: XAC mm HAM SO BAC NHAT VA SU

TUONG GIAO GIUA DO THI CAC HAM SO

Phuong phdp giai a i ,

• De xac djnh ham so bac nha't ta la nhu sau Gpi ham so can tim la y = ax + b, a ;^ 0 Can cii thee gia thiet bai toan de

thiet lap va giai h§ phuong trinh voi an a, h, tu do suy ra ham so can tim

• Cho hai duong thing dj : y = a^x + bj va d j : y = a2X + bj • Khi do:

^) dj va d2 trimg nhau <=> ai =a2

bi = b , ' ,

aj =32 b) d , va d2 song song nhau <=> s, , ;

b) d d i qua C(3;-2) va song song voi A : 3 x - 2 y + l = 0 _ j ( 0 <

c) d di qua M ( l ; 2 ) va cM hai tia Ox, Oy tai P, Q sao cho S^^QVQ " h o nhat

d) d d i qua N (2; -1) va d i d ' voi d ' : y = 4x + 3

Lai gidi

Goi ham so can t i m la y = ax + b, a 0

a) Vi A e d va B e d nen ta co he phuong trinh

Vay ham so can t i m la y = -4x + 7

1 1

Va d 1 d ' => 4.a = - 1 o a = -— thay vao (4) ta dup'c b = - — •

Vay ham so can t i m la y ~ ~ ^ ^ ~ ^ •

V i du 2: Cho hai d u o n g thang d : y = x + 2m, d ' : y = 3x + 2 ( m la tham so)

a) Chung m i n h rang hai d u o n g thang d, d ' cat nhau va tim toa do giao diem ciia chiing

b) Tim m de ba d u o n g thang d, d ' va d " : y = - m x + 2 phan biet dong quy

Lai gidi ,^ ,

a) Ta CO a^^ = 1 ^ a^j = 3 suy ra hai d u o n g thang d, d ' Ccit nhau

Toa do giao diem ciia hai d u o n g thang d, d ' la nghiem ciia he p h u o n g trinh

y = x + 2m [ x = m - l

- ^ suy ra d, d ' cat nhau tai M ( m - l ; 3 m - 1)

y = 3x + 2 [y = 3 m - l ^ ' b) Vi ba d u o n g thang d, d', d " dong quy nen M e d " ta co

m = l ' =

m = -3 ' ' '

• V a i m = l ta CO ba d u o n g thang la d : y = x + 2, d ' : y = 3x + 2, d " : y = - x + 2,

phan biet va dong quy tai M ( 0 ; 2 )

• Voi m = -3 ta CO d ' = d " suy ra m = -3 khong thoa man ''''

m = 1 la gia trj can t i m ' 3m - 1 = - m ( m - 1 ) + 2 <=> m^ + 2m - 3 = 0 <o

V i d u 3: Cho d u o n g thang d : y = ( m - l ) x + m va d ' : y = |m^ - i j x + 6

^) Tim m de hai d u o n g thang d, d ' song song voi nhau b) T i m m de d u o n g thSng d cMt true tung tai A , d ' cat tryc hoanh tai B - ,^ggo_cho tam giac O A B can tai O

Lai gidi

Voi m = 1 ta CO d : y = 1, d ' : y = 6 do do hai d u o n g thAng nay song song voi

Phdn loai va phuomg phdp gidi Dai so 10

Voi m = - 1 ta CO d : y = -2x - 1 , d ' : y = 6 suy ra hai d u o n g t h i n g nay cat

Doi chieu v o i dieu kien m ±1 suy ra m = 0

Vay m = 0 va m = 1 la gia trj can tim

b) Ta CO tpa dp diem A la nghi^m cua he

m = : - 2

m = 2 (thoa man) Vay m = +2 la gia t r i can t i m

Ca 2 B A I T A P LUVeN T A P

Bai 2,13: Cho ham so bac nhat c6 do thj la duong thang d T i m ham so do biet:

a) d d i q u a A ( l ; l ) , B(3;-2) , v i , , ,

b) d d i qua C(2; -2) va song song voi A : x - y + l = 0

c) d d i qua M ( l ; 2 ) va cat hai tia Ox, Oy tai P, Q sao cho AOPQ can tai O

d) d d i qua N ( l ; - l ) v a d i d ' voi d ' : y = - x + 3

Huang dan gidi

Gpi ham so can t i m la y = ax + b,a 0

a) V i A 6 d va B 6 d nen ta c6 h ^ phuong trinh

l = a + b

3 - - 2 a + b

2 ' = " 3 2 5 =>y = — x + -

a < 0, b > 0 ,

va c i t Oy tai Q(0;b) v o i

Taco OP = O Q c ^ - - - b < : : > b ( a + l ) = 0 « b = 0 (loai) a = - l

T a c 6 M e d = > 2 = a + b = > b = 3 1 Vay ham so can t i m la y = - x + 3 " £

d) D u o n g t h i n g d di qua N ( l ; - 1 ) nen - l = a + b ^ ^'pi

Va d l d ' = i > a = l suy ra b = -2 \^„^

Vay ham so'can t i m la y = x - 2 •sr.+ | c

Bai 2.14: T i m m d e b a d u o n g thang I

d : y = 2x, d': y = - x + 6, d": y = m^x + 5m + 3 phanbiet dong quy

Huang dan gidi " •'

Tpa dp giao diem(ne'u c6) cua hai d u o n g t h i n g d , d ' l a nghiem ciia hf

phuong trinh y = 2x x = 2 suy ra d, d ' cat nhau tai M ( 2 ; 4)

Lai gidi

a) TXD: D = R , a = 3 > 0 suy ra ham so dong bien tren

Bang bien thien

b) TXD: D = R , a = - - < 0 suy ra ham so nghjch bien tren

Bang bien thien

p u o n g thang y = _x - 3, y = - 2 c5t nhau tai A ' ( - 1 ; - 2 )

V i du 2 Cho cac ham so': y = 2x - 3, y = - x - 3, y = - 2

a) Ve do thj cac ham so tren '

b) Dua vao do thj hay xac djnh giao diem ciia cac do thj ham so do

v a i true hoanh va cat true tung

tai diem c6 tung do b i n g -2

b) D u o n g thang y = 2x-3, y = - x - 3

cat nhau tai A ( 0 ; - 3 )

ang thSng y = 2x - 3, y = - 2 Ccit nhau tai A " i ; - 2 2

•;jr^^;^^;^S^d6\h\ham so CO do thi (C) (hinh ve)

a) Hay lap bang bieh thien cua ham so tren [-3; 3]

ia tri Ian nhat va nho nhat ciia ham so tren [-4; 2]

Phdtt loai va phuang phdp gidi Dai so 10

Do thi ham SO y = x + 2 d i qua A ' ( 0 ; 2 ) , B'(-2;0) Ky^,

Do thi ham so' y = — d i qua

M

* 2

2,

va song song v o i true hoanh

b) Giao diem cua hai do thi ham PO , ^ j ;

y = -2x + 3, y = X + 2 la M l

3 ' 3

O

3 ^ 3 3 ^ Giao diem cua hai d o thj ham so y = - 2 x + 3, y = - la M2

Ve do thj ( C ) ciia ham so y = |ax + b| ta lam n h u sau

Cdch 1: Ve ( C j ) la d u o n g thang y = ax + b v o i phan d o thj sao cho hoanh

dp X thoa man x > - - , ve (C2) la d u o n g thMng y = - a x - b lay phan do thi

sao cho X < - - K h i d o ( C ) la hop ciia hai do thj ( C j ) va (C2)

a /

Cdch 2: Ve d u o n g t h i n g y = ax + b va y = -ax - b roi xoa d i phan d u o n g

thang nam d u d i tryc hoanh Phan d u o n g t h i n g nam tren true hoanh chinh

l a ( C )

CM y:

• Biet truoc do thj ( C ) : y = f (x) khi do do thj ( C j ) : y - f (|x|) la gom phan :

- Giir nguyen do thj ( C ) 6 ben phai true tung;

- Lay doi x u n g do thj ( C ) 6 ben phai true tung qua true tung

• Biet truoc do thj ( C ) : y = f (x) khi do do thj ( C j ) : y = |f (x)| la gom phan:

- G i u nguyen do thi ( C ) 6 phia tren true hoanh

- Lay dol x u n g d o thj ( C ) 6 tren d u a l true hoanh va lay doi x u n g qua true

Voi x < 0 d o thj h a m so y = - x la phan d u o n g thang d i qua hai diem

B ( - l ; l ) , C ( - 2 ; 2 ) nam ben trai ciia d u o n g thang x = 0 b) Ve hai d u o n g t h i n g y = - 3 x + 3 va y = 3x - 3 va lay phan d u o n g t h i n g nam tren true hoanh

V i dv 2: Ve do t h i ciia cac ham so sau

a) y = ix|-2 b ) y = | x | - 2

Lot gidi a) Cach 1: Ta eo y = x - 2 k h i x > 0

- x - 2 khi X < 0

Ve d u o n g t h i n g y = x - 2 d i qua hai diem A(0; -2), B(2; 0)

Va lay phan d u o n g thang ben phai ciia true tung

d u o n g thang y = - x - 2 qua hai diem A(0; -2), C(-2; 0) va lay phan d u o n g thang ben trai ciia true tung

Cdch 2: D u o n g t h i n g d : y = x - 2

f i q u a A ( 0 ; - 2 ) , B ( 2 ; 0 )

Phiin loai vn phucnig pltdpgidi Dai so 10

K h i do do thj ciia ham so y = |x| - 2 ia

phan d u o n g thang d nam ben phai ciia

I ' true tung va phan doi x u n g cua no qua

1 true tung

b) D o thi y = |x| - 2| ia gom phan:

- GiO nguyen do thj ham so

y = |x| - 2 6 phia tren true

D u a vao bang bien thien ta c6

m a x y = 5 khi va chi khi X = - 2

m i n y = l khi va chi khi X E [ 0 ; 1 ]

«««— [-2;2]

f l k h i x > - l b) T a c o y - | x + 2|-|x + l| =

Bang bien thien

2x + 3 khi - 2 < X < - 1

-1 khi X < -2

Cty TNim MTV DWU Khang Vift

Taco y ( - 2 ) = - L y ( 2 ) = l , , „ Dua vao bang bien thien ta eo

m i n V = - 1 khi va chi khi x < -2 max y = 1 k h i va chi khi x > - 1

p 2 BAI T A P L U Y E N T A P 7' Bai 2.17: Ve do thj ham so y = 2x - 3 Tir do suy ra do thi ciia:

( C , ) : y = 2|x|-3, ( C 2 ) : y = |2x-3|, (C3): y = |2|x| - 3|

Huang dan gidi

D o t h j h a m s o y = 2 x - 3 d i q u a A ( 0 ; - 3 ) , B ( 2 ; l ) t a g o i l a (C) (, •

• Khi do do thj ham so ( C i ) : y = 2|x|-3 la phan duge xac d j n h n h u sau

Ta g i i i nguyen do thj (C) 6 ben phai true tung; lay doi xirng do thj (C) 6 phan ben phai true tung qua true tung

• (C2): y = |2x - 3| la phan do thj (C) nam phai tren true hoanh va do thj lay doi xung qua true hoanh ciia phan nSm tren true hoanh ciia ( C )

• ( C 3 ) : y = |2|x|-3| la phan do thj ( C i ) nam phai tren true hoanh va do thj lay doi x u n g qua true hoanh eiia phan nSm tren true hoanh eiia ( C j )

Phdtt loai vd phuomg phdp giai Dai so 10

Huang dan giai

Taco y = |x-2|-3|x-l|

[-2x + l Khi x > 2 < -/'xi,.'

' ' = i - 4 x + 5 Khi l < x < 2 ,^

2 x - l Khi x < l Bang bien thien

x + 2 b) Bien luan so'giao diem cua do thj ham so tren vai dirong thang y = m theo m

Humg dan giai

diem ciia no vai duong thang y = m nhu sau:

Voi m > 1 thi c6 1 giao diem ,

Vai m = 1 thi c6 hai giao diem

Voi m < 1 thi CO ba giao diem '' * '

DANG TOAN 4: UNG DUNG CUA HAM SO BAC NHAT TRONG CHUNG MINH BAT DANG THUC VA TIM GIA TRI NHO NHAT, LON NHAT

Tim m de gia trj Ion nhat cua f (x) tren [1; 2] dat gia tri nho nhat

Lot giai

Dua vao cac nhan xet tren ta thay maxf(x) chi c6 the dat dupe t^i x = l

[l/*2] , J „ ^

hoac x = 2 Nhu vay neu dat M = max f(x) thi M > f (1) = |2 - m| va M > f (2) = |4 - m|

Khi do ham so dupe viet lai la y = |t - 3m + 4| voi t € [0;1] suy ra

A = maxlt - 3m + 4| = maxfl -3m + 4U5 - 3m +1) > I 3m + 4| + |5-3m|

|0,ii' I II M li 2

A p d u n g BDT trj tuyet doi ta c6

|-3m + 4| + |5 - 3m| = |3m -4| + 1 5 - 3m| > 1 " '

1 3 ' 3

Do do Dang thuc xay ra = ~ • & ^ caVi t i m la = " •

V i du 3: Cho a,b,c thupc [ 0 ; 2 ] « "^(^ 8"*^'" " ^ f f

C h i i n g m i n h rang: 2(a + b + c) - ( a b + bc + ca) < 4

Viet ba't dcing thuc lai thanh (2 - b - c)a + 2(b + c) - be - 4 < 0

Xet ham so bac nhat f (a) = (2 - b - c)a + 2(b + c) - be - 4 v 6 i an a G [ 0 ; 2 ]

Taco: f (0) = 2 ( b + c) - be - 4 = - ( 2 - b)(2 - c) < 0

f(2) = ( 2 - b - c ) 2 + 2(b + c ) - b c - 4 = - b c < 0

Suy ra f (a) < max (f (0); f (2)J < 0 dpcm

V i du 4: Cho cac so thuc khong am x, y, z thoa man x + y + z = 3

C h u n g m i n h rang x^ + y^ + z^ + xyz > 4 '

Lax giai

Bat dang thuc t u o n g d u a n g voi (y + z)^ - 2yz + x^ + xyz > 4

c:>(3-x)^ + x^ + y z ( x - 2 ) - 4 > 0 o yz{x - 2) + 2x^ - 6x + 5 > 0 (*)

Dat t = yz , do yz > 0 va yz < ''y + z^ ( 3 - x ) nen t e 0; ( 3 - x r

K h i do V T H la ham so bac nhat cua bien t, f(t) = (x - 2)t + 2x^ - 6x + 5

De chung m i n h ba't dang thuc (*) ta se chung m i n h f ( 0 ) > 0 va

nen bat dang thuc duge chung m i n h

Dang thuc xay ra <=> x = y = z = 1

Mat khac ta lai CO , '

xy + yz + zx - 2xyz S — o f(yz) = (1 - 2x)yz + x(l - x) - — < 0 (2)

Khi do ta thay rang

Neu X = i khi do BDT (2) thanh - ~ < 0 (hien nhien d u n g )

Ncu x ^ l thi f(yz) la ham so bac nhat Do do de c h u n g m i n h f ( y z ) < 0 ta

f f ( 0 ) < 0 chi can chung m i n h ( 1 - x ) ^

Vay la trong hai t r u o n g hop ta ket luan f(yz) < 0 Ta da giai xong bai toan

^ai 2.21: Cho • ' ' ' J ' ° 3 • C h u n g minh x^ + y^ + z^ + xyz > 4 ^ ' '

Hiiattg datt giai

gia thiet ta c6 x, y, z e [O; 3] va y

^ a t khac ta thay

, (y + zf {3-xf

va vz < = ;; •

4 4 + y H-z"^ + x y z > 4

^ x2 + (y + _ 2yz + xyz - 4 > 0 o x^ + (3 - x)^ - 2yz + xyz - 4 > 0

^ f ( y z ) = ( x - 2 ) y z + 2 x 2 - 6 x + 5 > 0 (3)

•^^'u x = 2 thi BDT (3) se thanh 1 > 0 (hien nhien dung)

7Q