Rèn luyện Toán nâng cao Đại số 9 - Nguyễn Cam

Hdn nffa vdi diTdng thing (d) thoa tinh chat cac giao di<^m ciia cac dufdng thing trong A^QOS ci cung mot nufa mat phing cd bd la (d) (luc nao trong Ajoojta ctlng chpn difdc mot d[r]

TS N G U Y E N CAM (Chu bien)

^ ThS N G U Y E N V A N PHUCfC

REN LUYEN TOAN NANG CAO

* Phan loai cac dang toan nang cao dinh cho hoc sinh khi gi6i

* Tuyen chon cac de thi tuyen sinh Idp 10 vao cac triTdng chuyen

THLT VIENTIfvHBINH THUAN

LdlNOIDAU

Cung vdi viec bien soan bo sach Ren luy^n todn can ban LOP 9, chiing toi gidi tliieu cuon Ren luyen todn ndng cao DAI SO 9 nham giup

cac em hoc sinh nang cao trinh do de tham gia cac ki thi tuyen vao Idp 10

thuQC cac trirdng chuyen, Idp chuyen

Cuon sach bao gom 9 chifcfng :

ChUdng 1 : Dang thiJc

ChiTdng 2 : Bat dang thiJc

Chufdng 3 : So hoc

ChiTdng 4 : Gia tri Idn nhat va gia tri nho nhat cua ham so

ChiTdng 5 : Phifdng trinh

ChUdng 6 : He phifdng trinh

ChiTdng 7 : Do thi ham so

ChiTcfng 8 : Mot so bai toan ve td hdp ; dpng tiJ

ChiTdng 9 : Mot so de thi vao Idp 10

Vdi muc dich h5 trd dac life viec tuT hoc ciia minh, cuon sach diTc^c

trinh bay theo bo cue nhiT sau : md dau mSi chufdng la phan torn t^t cac kien thilc can ye'u ma hoc sinh phai nam vffng; tiep theo la mot he tho'ng b^i tap CO hiTdng din each giai diTdc sap xep theo thiJ tif kho dan

De viec hoc cd ket qua to't, sau khi doc va tim hieu that ky cac bai tap CO hiTdng din hoc sinh nen co gang tif minh giai cac bai tap ay mot cdch doc lap cho den khi thuan thuc Vdi each lam ay, hy vong cac em se tif minh kham pha ra nhieu each giai khac mot each thu vi

Raft mong nhan diJdc cac y kien phe binh cua quy doc gia de cuon

sich ngay cang diTdc hoan thien va phuc vu ban doc tot hdn

TM.nhdm tdc gid

TS Nguyen Cam

3

3) {a + bf =a' -h^a'b + ^ab^ +b\

4) {a-bf=a' -3a'b +3ah^-b\

1.1: Tinh cac bieu thitc :

- 2 -3 2) Khi a = - T a c 6 : M = — ^

1) A = ^Ja^ -6a + 9 =^(a-3y =\a-3\

a-3 neu a-3>0 f a - 3 new a > 3

3 - a n ^ M a - 3 < 0 3 - a neu a<3

2) 5 = V7 + V x ' - 4 x + 4 = V7 + ^ ( x - 2 ) '

= | x | + | x - 2 |

-x-x + 2neu x<0 {-2x + 2 ne'ux<0 x-x + 2 neu0<x<2 = l2 neu 0<x<0

1) T i m dieu kien cua a de M CO nghia

2) Rut gon bieu thu-c M 3) V d i gia tri nguyen nao cua a thi M c6 gia tri nguyen ?

1: l + \/aj (a + l)(x/a-l)

a>0

1) Bieu thu'c M c6 nghia <=> \ 4a^0 \a>0 <=> {

2) Vdi dieu kien a > 0 va \a c6:

' 1 ^ ( v / ^ - l j (l + V^)(-v/^-lj

a > 0

1: 1 + VaJ (a + l)(Va-l) (a + l)(Va-l)

a + \ a + 1 3) Ta c6: M = -^-^ = 1 - — B i e u thu'c M nhan gia tri nguyen

a + 1 a + 1 khi

Vay khi a = 0 thi bieu thu'c M nhan gia tri nguyen

1.8: Cho bieu thu'c:

M =

b'

vdi |a| > |^>| > 0

Rut gon bieu thu'c M

Vdi dieu kien a\>\b\> 0 Tacd:

10

a' + l a ^ l a ^ + a^-b^-a^+ la^a"-b"-a" + b" 4|a|Va^-6'

a^-[a'-b') b'

'A\a \4^^ 4\a\b^yja^-b^ [-1 neu a<Q

DC 1.9: Riit gon bieu thu'c:

3 (yfai) - b) (yfa - f + 2av'a + b ^

1.10: Rut gon bieu thrfc sau:

yfx^^l

2) Ta cu :

J : = ^ / x - l - 2 ^ / ] ^ ^ + ^/x + 7 - 6 V x ^

c - a = 0

Vay, nSu a^ +b^ -ab-bc-ac = 0 thi a = b = c

)e 1.17: Cho ba so' a, b, c sao cho c ;t b, c ?t a + b va

nen: = a^ + {c^ + lab ~ lac - 2bc)

s = (a^ - 2ac + ) + 2Z)(a - c)

= {a-cf+lb{a-c) = {a-c){a-c + lb)

Wngtuf: b'^ =b^+(c^+lab-lac-lbc^

= [b^ -lbc + c^) + la(b-c)

= (b-cy +laib-c) = {b-c)(b-c + la)

a' + {a-cf _ (a - c){a-c + 2A)^+ (a - cf

Do do:

b'+ib-cf ~ ib-c)(b - c + la) + {b- cf

(a-c)(la-lc + lb) _a-c

~ (b-c)(2b-lc + la) b-c (vi a+b^c)

D6 1.18: Cho a, b, c doi mot khac nhau va thoa

(b-cf {c-af (c-bf

De 1.19: Cho ba so' a, b, c thoa a + b + c = abc

Chu-ng minh: a(A' - l)(c' -1) + b(a^ - l)(c' -1) + c(a' - \)(b^ -1) = 4abc

GiSi

Ta CO: a(b^ - l)(c' -1) + b(a^ - l)(c' -1) + c{a^ - \)(b^ -1)

= ia + b + c) + (ab^c^ + baV + ca^b^) - (ab^ + ba^ +ca^+c^a+be" + b'^c)

= abc + {ab^c^ + ba^c^ + ca^b'^) - [ab{b + a)+ ac{a + c) + bc{b + c)

= abc + (ab^c^ + ba^c^ + ca^b^) - [ab(abc -c)+ac{abc -b) + bc{abc - a)

(vi a + b + c = abc nen a+b-abc-c,b + c-abc-a,a + c = abc-b)

= abc + (ab^c^ + ba^c" + cc^b^) - (a^Z»'c+a'c^i+ab'^c^)+l>abc = 4abc

Vay ddng thiJc da diTdc chiJag minh

1.20: Cho - = ^ = - = wa b c (m>0). Chiang minh ring :

nen phai xay ra b'=b'

\6i =a^ => a^(a-l)-0=> a = 0 hay a-l=>a^=a

vadodd S = a + b^+c^ =a^ +b^ = \ VayS = l

a>b <:i> a-b >Q

a>b <^ a-b>0

Chu v; Vdi moi so' thrfc X, ta cd: > 0

II C A C TINH C H X T C d BAN CUA BAT DANG THLfC

<:>a\a-b)-b\a-b)>0<>(a^-b^)(a-b)>0 o(a + b)(a-b)(a-b)>0<:>ia + b)(a-by >0 (*)

V i a > 0 , 6 > 0 nen a + b>0 va (a-bf >0 n^n (*) diing

Vay +b^ >a^b + ab^

Bi 2.2: Cho a > 0, b > 0 Chitng minh: + > ^/a + V ^

Vay: - ^ + - ^ > y/a + yfb

f)i 2.4: Chitng minh cac ba't dang thtfc:

1) a " * > a^6 + aZ>^ vdi moi a, b 2) a ' + 6 ' > - v d i a + 6 > 1

D6 2.6: Chu'ng minh rang vdi m o i a, b, c, d, e ta c6:

(a^ >a{b + c + d + e)

a ,2 •

a* b^

T a c d : a ' < — + — •

b' a'

O a^t' + a'b' <a'+b'<^a'- a'b' + b'- a'b' > 0

O a'(a' - fc') - b\a^ -b')>Oo(a'- b')ia' -b')>0

o(a' -b')ia' -b'){a' +a'b'+b')>0

<:>ia'-b')\a'+a^b^ + b')>0 (dilng)

V a y : a +b' <^ + ^

b a

6 ^ 6

D d 2.8: C h o a.b>0 Chu'ng minh rang: (a + bfxy<{ax + by)(bx + ay)

T a c(5: (a + b)"" xy<(ax + by)(bx + aj)

<=> a^xy + 2aZ7jc>' + b^xy < abx^ + aby^ + c^xy + b^xy

<:> abx^ + aby'' - 2aZ?x>' > 0 <::> ab{x^ + -2^:^) > 0

<^ab{x-yf>Q

Ba't d^ng thi?c n a y diing vi ab > 0,(jc - y)^ > 0

V a y : {a + bf xy <{ax-\- by)(bx + ay)

Bi 2.9: C h o a + > 2 Chu'ng m i n h rang: a^ +b' <a' +b\

Gidi

TsiCd: a'+b'>a'+b^

oa'-a'+b'-b'>0 c>a\a-l) + b\b-l)-{a-\)-ib-l) + a + b-2>0

Mat khac, ta lai cd: (1 + a + + c)^ > 0

^\ a^ +b^ + +2(ab + bc + ca) + 2(a + b + c)>0

=>2 + 2(ab + bc + ca) + 2(a + b + c)>0 (vi a^ + b^ =1)

Do do de chiing minh (1) ta chi can chiJng minh bat ddng thrfc sau:

a(l-b) + b(\-c) + c(\-a)<\)

That vay, ta c6: 1 - a > 0,1 - ^ > 0,1 - c >,a^c > 0

nen (1 - a)(l - ^)(1 - c) + a^c > 0

a + c

2c-2ac

a + c a^+3ac c'^+3ac 1

(1) 2.14: Cho ba s6' a, b, c thoa: ab + bc + ca>0 (2)

=>bc>-a(b + c)>0 (vi a<0,b + c>0)

Do a < 0 va b o 0 nen suy ra abc < 0 Dieu nay trai vdi gia- thi6'l

la abc > 0

Trirdng hop b < 0 hoac c < 0, ta ly luan trfdng tif

Vay ta da chiJng minh r^ng: a > 0, b > 0, c> 0

Bi 2.15: Cho 0 < a < l , 0 < b < l , 0 < c < l Chitng minh trong ba bat

dang thtJc sau cd it nha't mot bat dang thiJc la sai:

ail-b)>]-, b(l-c)>]-, c(\-a)>]- 4 4 4

Gi^i

Ta chiJng minh bang phan chiJng:

Gia su" ca ba bat ding thiJc deu dung, liic dd nhan ve'theo ve ta cd:

a(l-a)bil-b)c(\-c)>

^ 3

(1) Mat khac ta cd:

Dieu nay mau thuan vdi (1)

Vay ta phai cd: Trong ba bat dang thiJc trong di bai cd it nhat

mot bat dang thiJc sai

§ 2 BAT DANG THLfC COSI (Cauchy)

I BS't d a n g thufc C6si cho h a i so' k h o n g fim

C h o a > 0 , & > 0 :

Ta c6: a + b > 2 ^

Da'ii dang thiJc xay ra ^ a = b

I I Ba't dang thuTc C6si cho n s6' k h o n g &m

Cho n so'khong am: a^,a^,•••,«„ •

T a c 6 :

Dau dang thiJc xay ra <^ = = = a „

D I 2 I 6 : Cho a, b, c > 0 Chitng minh:

2) (a + fe + c) Da'u dang thiJc xay ra k h i nao ?

Da'u dang thiJc xay ra k h i nao ?

A p dung B D T Cosi cho hai so'khong am, ta c6:

D(J 2.18: Cho a > 0, b > 0 Chitng minh: (a + bf +

Dau dang thiJc xky ra khi nao ? -A \

Tir (3) va (4) suy ra: {a + b) +

Dau ding thtfc xay ra <^ a = ^ = 1

>8

2.19: Cho a,b, c > 0 va a + b + c = L Chtfng minh:

>64

a A c Dau ding thiJc xay ra khi nao ?

De 2.21: Cho a > b > 0 Chiang minh: a +

' DS'u dang thiJc xay ra khi nao ?

0, b > 0 Ap dung baft dang thiJc Cosi cho hai s6' du'dng, ta c6:

2 \fja4b <4a+^ =^2i/ab <4a + yfb

2^b-j^<(^a + S ) ^ ^ 4a^4b ^ '4a+4b

2 ^ 4a + y[b < '4ab (dpcm).i

D d 2 2 3 :

1) Cho a>\,b>\ Chtfng minh: a^Jb -X + b^a- \ 2) Cho ba s6' a, b, c doi mot khac nhau ChiJng minh:

{a-Vbf , ib^-cf {c + af {a-bf"^\b-cf • (c-a) 2 — >2

lb Ic la

a — b b — c c — a

Sabc ' ia-b){b-c)(c-a)'

(3) (1) +(2) + (3): — + — + — + — + — + — > 2 6 + 2a + 2c c a c b a b ~

> ^ (4),

(5) (6)

Tac6: 3a' + 3 ^ ' +Ab' >^l]3a\3b\Ab\

Mat khac: IJ^JA > ^ 3 3 3 = 3 (2)

(1) va (2) ^ 3a' + 7b' > 9ab^ (dpcm)

§ 3 BAT D A N G THUfC TRONG TAM G I A C

Cho tam giac ABC Dat: BC = a; AC = b; AB = c

Bi 2.26: Cho a, b, c la do dai canh ciia mot tam giic

ChiJng minh rkng: +b^ +c'^ < 2{ab + ba + ca)

GiSi

Theo tinh chat trong tam giac ta c6:

0 < a < + c < a(Z? + c)

Q><b<a^-c^b'^ <b{a^-c) Q<c<a^-b=>c^ <c{a^-b)

C6ng v6' theo ve ba ding thtfc tren ta dufdc:

a^ <2(a^ + fea + ca)

2.27: Cho a, b, c la do dai ba canh ciia mot tam giac

Chiang minhr^ng: abc>(a + b-c)(b + c-a)(a + c-b)

Giii

Tac6: a' >a^ -{b-cf ={a + b-c)(a + c-b)>0

39

b"" >b^ -(a-cf =(a + b-c)(b + c-a)>0 '^c^ -(a-bf =ia-\-c-b)ib + c-a)>0

Nhan ve' theo ve'ba bat dang thiJc tren ta cd:

a'feV >ia + b-c)\a + c-b)\b + c-af

=^ abc>ia + b- c){b + c - a){a -^c-b)

2.28:

1) Cho x > O v ^ y > O C h u ' n g m i n h r ^ n g : - + - > — ^ D i n g

X y x + y

thtJc xay ra khi nao ?

2) Trong tam giac ABC cd chu vi 2p = a + b + c (a, b, c la do dai

ba canh tam giac) ChiJng minh rang:

D i n g thu-c xay ra khi

De 2.29: Cho a, b, c la do dai ba canh cua mot tam giac vdi chu vi 2p

Chiang minh rang: (p - a){p - b){p -c)< abc

{p-a){p-b){p-c)^ {a + b-c) (b + c-a) (a + c-b)

2 2 2 Mat khac: (a + b-c)(b + c-a)(a + c-b)< abc.( De 2.27 )

Vay: {p - a){p - b){p - c) < abc

= 1 (dpcm)

§4 PHUdNG PHAP LAM TRQI

Dilng tinh chat cua bat dang thtfc de du'a bat dang thrfc can chiJng minh ve dang c6 the tinh diTdc theo phrfdng phap tinh tong hffu han

Gia su" can tinh tong: S„=U,+U^+U^+ + [/, + + 1 / „

Ta phan tich: — a,^ —

K h i d o :

S„ -a^) + (a^-a,) + (a,-a,) + + (a„ - ) = - a „ + , •

2.31: Cho n la so' nguyen thoa n > 1 ChiJng minh:

<^ 1 > 0 Die 11 nay luon diing

C6ng vd" theo v<S' n ba't ddng thitc tr^n ta duTdc:

2 2 2

5 + 2 5 + 4 ? ^ - + - ^ '

1 1 (2« + l ) ' ^ 2 2n + 2 ' ^ 2 '

CHlTOfNG III: SO HOC

I.TINHCHIA HEX

1) Dinh nghia: Cho a, b la cac so nguyen Ta noi a chia bet

cho b neu ton tai mot so nguyen q sao cho:

a = bq

Ki hieu: a:b Khi do ta cung noi b chia het a va ki hidu: b|a

2) Tinh chat:

a) Neu: a chia het cho b va b chia he't cho c

thi: a chia he't cho c

b) Ne'u: a, b cung chia he't cho c

thi: a + b va a - b chia he't cho c

c) Ne'u: a chia he't cho c hay b chia het cho c

thi: a b chia he't cho c

3) Dau hieu chia he't:

a) Dau hieu chia he't cho 2: chi? so tan cung la chan ;

b) Dau hieu chia he't cho 3 (cho 9): tong cac chi? so chia

he't cho 3 (cho 9)

c) Dau hieu chia he't cho 4 (cho 8): hai (ba) so' cuo'i chia

he't cho 4 (cho 8)

d) Dau hieu chia he't cho 5 : chu" so' tSn ciing 1^ 0 hoSc 5

e) Dau hieu chia he't cho 11 • hieu giffa tdng cac chiJ so d

vi tri chan va tdng cac chff so'd vi tri le chia he't cho

11

II SO NGUYEN TO

1 Mot so nguyen du'dng du'cJc ggi la so nguyen to ne'u no chi

chia he't cho 1 va cho chinh no

• Nhtf vay theo dinh nghia nay, so' 1 la so nguyen to

Tuy nhi^n trong cic s6' nguyen t<5' tham gia l^m thiYa

s6' ciia mot tich, ngtfdi ta qui rfdc khong ki d6'n so' 1

2 S6' nguyen dufdng kh6ng la s6' nguySn t6' duTcJc goi Ik hdp

III U(5C CHUNG L(5N NHAT - BOI CHUNG NH6 NHXT

1) Binh nghia 1: Udc chung Idn nhS't (l/CLN) ciia hai s6' a b khong ddng thdi bkng 0 la s6' nguySn Idn nhat chia ha't ca a

vkb

Kihi6u: (a ,b) d^ chi UCLN cila a vk b

2) Binh nghia 2: c^c s6' nguyen a, b diTdc goi Ik nguySn t6' cing nhau n^'u: (a, b) = 1

3) Tinh chat: ra b) = ra \a + h\\= (\a + h\h)

4) Cach tim UCLN cua a vk b:

• Phan tich a va b thknh thiTa s6' nguyen t6'

• Lay tich cac thu'a s6' chung vdi s6' mu be nhat

5) Binh nghia 3: Boi chung nhd nhat (BCNN) cua hai s6' a va b

la so' h6 nhat chia he't cho ca a va b

Kijiieu: [a;^] de chi BCNr cua a va b

6) Cich tim BCNN cua a va b:

• Phan tich a vk b thanh t'lii'a so' nguyen t6'

• Lay tich cac thil'a s6' chung va rieng vdi s6' mu Idn nha't

IV.DONGDUTHLTC

1) Binh nghia: Cho m la s6' tif nhi^n khac 0 Hai s6' a, b dtfdc goi la ddng dtf modun m, ne'u trong phdp chia a vk b cho m

ta duTdc cilng dif so'

Khi 66, ta k i hien: a = b (mod m)

2) a = b(mod m)-^ a — b + km vdi k la so' nguy6n

Khong the nho't 7 chit tho vao 3 chiec long, sao cho

moi long c6 khong qua hai chit tho

* Nhu" vay neu nho't 7 chit tho vao 3 chiec long thi it nha't phai

c6 mot chiec long c6 tu" 3 chu tho trd len

Trong do: n! = 1.2.3 n (vdi n6 iV*) va 0! = 1

3.2: Chifng minh rang tong 2p + 1 so tu" nhien lien tie'p chia he't

3.3: Cho n e N' Tim UCLN cua hai s6': 2n + 3 va n + 7

Gi§i

Ap dung tinh chat: (a,b)= {\a±b\,b)^{a,\a± b l)

Ta c6: (2n + 3;n + 7) = (l n - 4 l,n + 7) = (l n - 4 1,11)

Nhtfng 11 Ik so' nguyen to' nen so' can tim la 1 hoSc 11

Khi do: * Khi In - 41 la bpi so cvia 11 thi:

UCLN(2n + 3;n + 7) = 11

* Khi In - 41 khong la boi so' cua 11 thi:

UCLN(2n + 3;n + 7 ) = 1

3.4: Mot so' nguy6n du'Png N c6 dung 12 tfdc s6' (du'dng) khac

nhau ke ca chinh n6 va 1, nhiTng chi c6 ba tfdc so' nguyen t6' khac

nhau Gia su" tong cua cac ufdc so' nguyen to' la 20, tinh gia tri nho

nhat C O the cd ciia N

N nho nhat nen N - 2^5.13 = 260

3.5: Chitng minh r^ng c6 s6' nguyen drfdng chi chiJa cac chi? s6' 0

va chi? so 1 va so do chia het cho 1999

Xet 2000 so'nguyen du-png sau: 1; 11; 111; ; H L ^

2000 chuso'X

Cac s6'tr^n khi chia cho 1999, ton tai it nhat hai sd'chia cho

1999 cd cilng so'du" (nguyen ly Dirichlet) Goi 2 sd'dd la :

• Khi n = 1, A = 0: khdng la so' nguyen t6'

• Khi n = 2, A = 5: Ik so' nguyen t6'

• K h i n > 2 : T a c d : n - l > 2 va n ' + l > 1 0 n a n A l a

hdp s6'

3.8: Cho p va + 2 la hai s6' nguyen t6' Chitng minh: + 2

cung la s6' nguy6n to'

D i e u nay trai v d i gia thie't

V a y khong xay ra tru'dng hdp r = 1

• V d i r = 2 : T a c 6 / ? ' + 2 = 3 ( 3 m ' + 4 m ) + 4 + 2

= 3 ( 3 m ' + 4 m + 2)

- , V i : 3m^ + 4m + 2 > 1 nen + 2 la hdp so

D i e u nay trai v d i gia thie't

vay cung khong xay ra tru-dng hdp r = 2

T d m l a i : K h i p va /? V 2 la so' nguyen to t h i : p = 3

K h i d d : + 2 = 29 cung la so' nguyen to' (dpcm)

3.9: ChiJng minh rang:

1) V d i m o i so'nguyen du'dng n, phan so'sau la to'i gian: ^ ^ " ^ ^

V a y : 21n + 4 va 14n + 3 la hai so' nguyen to' cung nhau

td-c la: i l ^ i ± l la phan so t o i gian

14n + 3 2) Ta chiJng minh bang phan chiJng

G i a i su- rang V2 e g v d i ^ = ^ (viet du^di dang to'i gian nghia la

Ta dung phufdng phap qui nap de chiJng minh

• Vdi n = 1: De thay: 1376^ = 1376(mod 2000)

Ttfc menh dd dung khi n = 1

• Gia su" menh de ddng tdi n = k Ttfc la:

Nghia la mdnh de diing tdi n = k + 1

Vay vdi moi s6'nguyen difdng n, ta cd: 1376" = 1376(mod2000)

Bi 3.12: ChiJng minh r^ng s6' jc = 10^" +10'" +10^" +1 chia het

cho 7, 11 va 13 neu n le Tim so' dif cua phdp chia x Ian liTdt cho 7,

l l , 1 3 v a l l l

Giii

.x = 10'"+10'"+10'"+l

TH,: Vdinle.tacd: * 10'" +1 = (10')" -(-1)"

chia het cho 10'-(-1) va: 10'-(-1)-1001 =

7.11.13-Do d6: 10'" + 1 chia het cho 7 ; 11 ; 13

Do dd 10'" +1 chia cho 111 du" 2 vdi moi neN

Tu-dng ttf 10'" +1 chia cho 111 dtf 2 vdi moi n e iV

Suy ra x chia cho 111 dtf 4 vdi moi neN

Tom lai: x chia cho 7 ; 11 ; 13 duT 0 n^'u n le ; dtf 4 na'u n chan

X chia cho 111 du" 4 vdi moi neN

Bi 3.13: Cho so' nguyen n > 1 ChiJng minh ring: n" -«'+«-1 chia he't cho (n-lf

Giii

Nhan x^t rkng vdi n = 2 thi : n" -n^ +n-l= (n-\f= 1 nan bai

, toan hien nhian dung

Bay gid ta xet n > 2:

Do do tu" (1) ta s«y ra rang n" -n^ +n-\a het cho (n - 1 ) l

De^ 3.14: Xet ba so t\i nhien a, b, c th6a he thiJc: ^b^+c\

ChiJng minh r i n g neu a, b, c nguyen to' cung nhau thi a la so' le va

trong hai so' b, c c6 mot so' le va mot so' chfn

Giii

Gia su" a chfn

V i a, b, c nguyen to' cung nhau nen b, c le

• a chan chia he't cho 4

• h,c le =>b + c chia he't cho 2 va be khong chia he't cho 2

^ib + cf - 2bc khong chia he't cho 4

=^b'^ khong chia he't cho 4

Nhtfng theo giai thie't: = ZP^ + c^ V6 l i !

Vay a le ma a' = + =^ c6 mot sd' le va m6t so' chJn

^b,c CO mot so' le va mot s6' chfn (dpcm)

Y)i 1.15: Chu-ng minh rling tong cac binh phu-Ong cua ba so' nguyen

hen tie'p khong the la lap phu'dng cua mot so' nguyen dtfcfng

Giii

Tru'dc he't ta chitng minh neu n e Z thi:

• chia cho 3 dtf 0 hoac 1

* chia cho 9 dtf 0 hoac 1 hoac 8

That vay: Dat n = 3m + r{n,m eZ;r=- 0;±1)

* = (3m + rf - 9m' + 6mr + r '

r ' = 0 ;1 nen n'chiacho 3 du-Ohoac 1 (1)

• * ==(3m + r)^ = 2 7 m ^ + 2 7 m V + 9 m r ' + r \

= 0 ;1 ; - l nen chia cho 9 du" 0 hoac 1 hoac 8 (2)

* Goi ba so nguyen hen tie'p Ian lu-dt la a - 1 ; a ; a + 1

Ta CO (a - 1 ) ' + a' + (a + 1 ) ' = a' - 2a + 1 + a' + a' + 2a + 1

= 3 a ' + 2

Tu" (1) ta CO 3a' + 2 chia cho 9 drf 2 hoac 5 ma chia cho 9 du" 0

hoac 1 hoac 8 (tu" (2))

Do do 3a' + 2 khong the la lap phu'dng cua mot so'nguyan drfdng

3.16: ChiJng minh rang tich bo'n so'nguyen du-dng l i ^ n tie'p khong

the la so' chinh phu'dng

Giii

Giai siJ ton tai so' nguyen m de:

m ' = nin + l)(n + 2)(n + 3) <^ m ' = ( n ' + 3n)(n' + 3n + 2)

= k{k + 2) (vdi k = n ' + 3n ) Tacd: y t ' < ) t ( ^ + 2 ) < ( ^ + l ) '

Dieu nay mau thuan vdi k{k + 2) = m '

TO do suy ra dieu phai chifng minh

3.17: Tim so' nguyen n sao cho: n' + 2n' + 2n^ +n + l la mot so

chinh phu'dng

Giii

Theo de bai ta c6: n* + 2n^ + 2n^ + n + 1 = (vdi m e Z )

^ A{n' + In" + 2n^ + rt + 7) = Am"

, Vay cac s6'c^n tlm la: n = 2, n = -3

f)i 3.18: Tim ba so' t\I nhien m, n, k d6i mot khac nhau sao cho

J- + i + - la mot so' tu" nhien

m n

Ta CO the gia su" r^ng m < n < k

Dat / = — + - + - vdi i la s6' tu* nhian

* V d i i = 2 tl>i c6 I 2 < — =4> m < - =4> m = 1

m 2 ' L t i c d d d i l u kientrd thanh: 2 = 1 + - + ^ = ^ - + ^ = !

n k n k

M a m = l < n < k n e n n > 2 , A : > 3 - ^ - + | < ^ + ^ - 7 < l

n A: 2 3 6

Do d6 khong thoa dieu kien neu tren

Ket luan: m = 2, n = 3, k = 6 la ba so' can tim

3.19: T i m cac so' nguyen difclng m, n sao cho: 2m + 1 chia he't

cho n va 2n + 1 chia he't cho m

GiSi

Trirdc he't ta xet triTdng hdp \

Ta c6: m < 2n + 1 < 2m + 1 (vi 2n + 11a boi so' cua m)

2n + 1 = m hay 2n + \ 2m hay 2n + \ 3m

(vi 4m > 2m + 1 nen kh6ng the nhan 2n + 4 = 4 m , )

• 2n + 1 = m:

Ta c6: 2m + 1 = 2(2n + 1) + 1 = 4n + 3 chia he't cho n nen suy

ra 3 chia h6't cho n, v i the phai c6: n = 1 hoac n = 3

V d i m = 1 t h i n = 1

Do do ta thu du-dc: (n = 1, m = 3),(n = 3, m = 7), (m = n = 1) jsfhan xet rang vai tro ciia m va n nhu* nhau trong de bki, do vay khi

1 < m < « ta tim them du-dc: (m = 1, n = 3), (m = 3, n = 7)

D6 3.20: M o t so' chinh phu-dng cd dang abed Bie't ab-cd = l Hay

tim so' a'y

Dat abed = vdi a, b, c, d la cac chff s6' Ta phai cd a^O, ngoaira, do ab-cd = l n€n : e^O

T a c d : = 1 0 0 a i + ^ = 1 0 0 Q + l ) + ^ Suyra: - 1 0 0 = 1 0 1 ^ - ^ ( n + 1 0 ) ( « - 1 0 ) = 1 0 1 ^ V i n^cd bo'nchu'so'nen n < 100, do dd n + 10 = 101, hay n = 91 Vay 8281

Bi 3.21: T i m so' tu* nhien n = ab cd hai chu" so' sao cho:

Ta lai bi^'t r i n g (x + yf > 4xy

Ket ludn: n = 91, n = 13, n = 63 la cac so' can tim

CHL/CfNGIV: GIA T R I L 6 N NHAT

VA GIA T R I NHO NHAT CUA HAM SO

Neu CO hang so'M sao cho:

N6u CO hing s6'm sao cho:

I DINH NGHIA Cho ham so' y = f(x) x^c dinh vdi x e D

thi m la gia tri nho nha't (GTNN) cua f(x)

K i hieu: m = min f{x)

Ghi chu: Tap xac dinh D la tap cac gia tri x sao cho f(x)

CO nghia

II C A C H T I M G T L N VA GTNN CUA H A M S d

1) Loai 1: Dilng tinh chat: |A| > A Dau "=" xay ra

2) Loai 2: Gia su' A, B la cac hang so, B > 0 va g(x) > 0

Cho: f{x) = A^ B Khi do: * f(x) Idn nha't ^ g(x) nho nhat

* f(x) nho nhat ^ g(x) Idn nhS't

B Cho: f(x) = A-

g(x) Khi do: * f(x) Idn nhat ^ gix) Idn nhat

* f(x) nho nhat ^ g(x) nho nhat

3) Loai 3: Dung dieu kien c6 nghiem cua phtfOng trinh bac

hai

Ghi chu: Cho ham so y = f(x) xac dinh tren tap D

Neu ph^dng trinh y = f{x)c6 nghiem

xeD ^ a<y<b

thi min f(x) = a va max f{x)^b

4) Loai 4: Diang cac linh cha't cua bat ding thiJc

4.1: TimGTNN cua : j = yjx-l-lyf^^ + ^Jx + l-6y[7^

Tac6: y = y]x-\-2y/x-2 +^lx + l-6yf7^

p i 4.2: Cho M = yjx + 4sjx -4 + ^x-A^x-A Tim x de M nho

nhS't va xac dinh gia tri nho nha't cua M

Gi^i

Dau "=" xay ra <^

Tac6: M = ^jx + Asfx^ + yjx-Ayjx - A

^ M = 4{X-A) + A^X-A^A + SJ{X-A)-A4I^^A

<^ 4 < ;c < 8 (thoa dieu kien)

Vay khi 4 < A: < 8 thi M dat gia tri nho nhat va min M = 4

4.3: Tim GTNN cua ham so: y = + -4;c + 4

4.4: Tim GTNN cua bieu tMc: A =

Dd4.8: Tim GTNN cua ham so: y = -^'^^

; 2

Vay ta cd khi ;c = ± 1 thi f = - / - - i - = 0 <s^ y = -

• 2 2 ^ 4

3 Vay y dat gia tri nho nha't la: min y = —

Vay khi = - ( l ± thi y = 6 nen y dat gia tri nho nha't la: min y = 6

4.10: Cho y = (x l)(x 2)(x + 4)(x + 5) Tim x de y dat gid tri

nhd nha't Xdc dinh min y

71

Vay khi x = - ( - 3 ± thi y dat gia tri nho nhat min y = - 9

4.11: Cho hai so' thiTc x, y thoa dieu kien x^ =1 Tim gia tri

Idn nhat va gid tri nho nha't ciia x + y

x = y =

Vdi x + y >-42

Dau " = " xay ra x = y Vay X + y dat gia tri nho nhat la

Vay max y = 2 dat du-dc khi ;c = ±sl2

DC- 4.13: Cho y = ^Jx^ +x + l + yjx^ -x + \ Tim x de y dat gia tri

nho nha't Xac dinh min y

V i x^+x + l = x' -x + l^

D d 4.14: Cho y = — T i m x de y dat gia t r i Idn nhat Xac dinh

X + 1 max y

Dat diTdc khi: x = y = —

Dau " = " xay ra ^ X = y - - Vay A dat gia tri nho nhat la 11

De 4.19: Tim gia tri Idn nha't ciia cac bieu thitc sau:

Vay khi X = y = 2 thi N dat gia tri Idn nha't la max N = 4

3) Dieu kien xac dinh la: x > l , y > 2,z > 3

y^2 + (y-l)>2^2(y-2)^^l^-^<~

I 2 1

• =:3 + ( z - 3 ) > 2 V 3 ( z - 3 ) ^ ^ ^ ^ < ^

(2) (3)

Khi X = 2, y = 4, z = 6 thi (1), (2), (3) trd thanh dang thiJc Do d6

1 44 1

nen maxM = 69-66.—= — itng vdi mmz = —

min M = 69 - 66 = 3 tfiig vdi max z = 1

78

§1 PHaONGTRlNH BAG HAI - PHl/dNGTRlNH BAG BA

DINH LY VIET

I PHl/dNG TRINH BAC HAI ax^+bx + c = 0 {a^O)

Lap biet thitc: A = 6^ - 4ac

1) Neu A < 0 thi phifcfng trinh v6 nghiem

2) Neu A = 0 thi phiTdng trinh cd nghiem kdp:

X, =x^= — la

3) Nd'u A > 0 thi phifcfng trinh cd hai nghiem phan biet:

la

Trirdng hdp dac biet:

• Neu: a + + c = 0 thi phu^dng trinh cd nghiem :

phUcfn I tilnh bac hai

2) N6u phu-dng trinh: ax^ + fejc + c = 0 cd hai nghiem

x^;x^ {x^ < j:2)thi: a > 0 | : * ax^+bx + c<0<^ x^ <x<x^

X < X

* ax'' +bx + c>0^

79