Tuyển chọn và hướng dẫn giải 500 bài tập Toán 10: Phần 1

Phần 1 tài liệu Tuyển chọn 500 bài tập Toán 10 giới thiệu các bài tập theo chuyên đề thuộc phần Đại số bao gồm: Tập hợp và hàm số, phương trình và bất phương trình bậc nhất một ẩn, hệ phương trình bậc nhất hai ẩn, bất đẳng thức,... Mời các bạn cùng tham khảo nội dung chi tiết.

P=^7 M A U T H O N G - L E M A U T H A O

500 bai t$p TOAlV 10

THLf VIEW TJNH biNH THUAN

N H A X U A T B A N H A N Q I

LCfl N O I D A I I

500 b a i t o a n loTp 10 dUcrc s o a n t h e o c a c y e u c a u :

B a m sdt s a c h giao k h o a h i e n h ^ n h

Cdc k i e n thufc ca hAn difcfc tap t r u n g vao tifng c h u y e n de

Trong moi chuyen de, chiing toi t o m li/cfc phan h' thuyet, chi r a c&ch

van d u n g l i thuyet vao cAc bai tap tis d& den k h o , n a n g cao d a n theo

htfdng tiep can miJc do cac de t h i tuyen sinh Dai hoc, t r i c h d a n v a

giai mot so de t h i tuyen sinh Dai hoc cac n a m 2002, 2003, 2004

D a c bi^t, trUdc m o i b a i g i a i , chiing toi dua ra p h a n hiTdng dSn, p h a n

n a y h e t siJc quan trong, no giiip b a n doc t i m duoc hiidng giai quyet

b ^ i t o a n , torn t ^ t qua t r i n h giai, de k h i giai xong b a n chi c a n n^m

difoc c a c y c h i n h la dii

Sdch dugc chia lam hai phan :

Phan 1 : D a i so (300 bai)

Phan 2 : H i n h hoc (200 bki)

Sau m o i p h a n , chiing toi c6 p h a n toan tdng hop n h ^ m giiip 'hoc s i n h

on l a i cac k i e n thiJc ccf b a n , tap t r a Idi cac cau h o i tr^c n g h i e m

D&t bi^t : Cuoi s a c h c6 10 de k i e m t r a n h k m giiip hoc s i n h t\i d a n h

gid v a k i e m t r a l a i cac k i e n thiJc d a hoc trong moi hoc k i

M6i de kiem tra c6 :

Cau h o i li thuyet

Cau hoi trftc n g h i e m

Toan t u luan

M a c dii da CO gSng n h i e u k h i b i e n s o a n , nhitog c h ^ c k h o n g t r a n h

k h o i t h i e u sot, chiing toi mong n h a n dufoc sU dong gop y k i e n xay

diTng ciia quy dong nghiep, cac e m hoc s i n h de Ian t a i b a n sau, s a c h

Chuyen de 6 Phifong t r i n h b§c hai mot an 66 Chuyen de 7 He phuong t r i n h bac hai doi xufng doi vdi x va y 78

Chuyen de 8 He phuong t r i n h d i n g cap bac hai 84 Chuyen de 9 Dau ciia t a m thiic bac hai 89 Chuyen de 10 Bat phiTOng t r i n h bac hai 98 Chuyen de 11 Xac dinh gia t r i ciia t h a m so m de tam thilc bac hai

CO dau khong thay d6i 109 Chuyen de 12 So sanh so a vdi hai nghiem ciia phuong t r i n h bac hai 113

Chuyen de 13 Phirorng t r i n h va bat phuong t r i n h c6 chiJa gia t r i tuyet doi

129 Chuyen de 14 Phifcfng t r i n h v^ bat phucrng t r i n h c6 chufa cSn thufc 137 Chuyen de 15 Tdng hop cudi nam 153

^ h ^ n HiNMHOC

Chuyen de 1 Vecto • 172 Chuyen de 2 True - Toa dp t r e n true 188 Chuyen de 3 He true toa dp Descartes vudng goc 194

Chuyen de 4 T i so luong giac 209 Chuyen de 5 Tich v6 hudng ciia hai vectP 222

Chuyen de 6 He thiic lupng trong t a m giac 235 Chuyen de 7 Giai t a m giac - f n g dung thifc te 244 Chuyen de 8 He thiJc lUcJng trong di/dng t r o n 250 Chuyen de 9 Tdng hop cudi nSm 256

DQ k i e m t r a 290

De k i e m t r a Hoc k i I 290

De k i e m t r a Hoc k i I I va cuoi n a m

3

c i H y 6 n 1 T6P HOP V6 HflM so

Kien thurc ca ban

I) Tap x a c djnh c u a ham s o y = f(x)

• Tap xac dinh ciia h a m so' y = f(x) la : D = |x e R \ f( x ) c 6 n g h i a )

• Cach t i m tap xac dinh ciia h a m so'y = f(x)

BU<Jc 1: T i m d i l u kien ciia x (x e R) de f(x) c6 nghIa

(f(x) duac xac dinh)

• BtfdTc 2 : Viet dieu kien t r e n difdi h i n h thdc tap hop

Q(x) > 0

II) Ti'nh ddn dieu c u a ham s o

Cho h a m so' y = f(x) xac dinh t r e n khoang (a; b) va X i , X2 e (a; b)

H a m so' y = f(x) dong bien tren H a m so' y = f(x) nghich bien tren khoang (a; b) k h i : khoang (a; b) k h i :

X ] < X 2 = > f(X]) < fiXa) X] < X 2 f(Xi) > f(X2)

H a m so' y = f(x) don dieu tren khoang (a; b) k h i va chi k h i ham so

y = f(x) dong bien hoac nghich bien t r e n khoang (a; b)

C a c h khao sat tinh ddn dieu cua ham so

• Bi^drc 1: T i m tap xac dinh D, lay x,, X 2 £ D, t i n h y j = f(xi) va y 2 = f(x2)

• Elide 2 : T i n h y i - y 2 theo x i va X2

• BUoTc 3 : Cho x i < X2 <=> X] - X 2 < 0

• Neu y i - y 2 < 0 hay y i < y 2 : H a m so dong bien t r e n khoang (a; b)

• Neu y i - y 2 > 0 hay y i > y 2 : H a m so nghich bien tren khoang (a; b)

5

ni) T i n h c h i n 1^ c u a mQt h a m s o

Gia sii h a m so y = f(x) c6 tap xac dinh D thoa man d i l u kien

V x , x e D = > - x e D (Ta bao D la tap doi xiing qua O)

\

y

O

a < 0 [b = 0

b) Neu a = 0 thi y = b la h a m so' h k n g v a do t h i la dudng t h i n g eung

phLfOng vdi true h o a n h Ox (Ta bao h a m so' h i n g c6 do t h i l a dudng

• Dong bie'n t r e n khoang (0; + » )

• Nghieh bie'n t r e n khoang (- oo; 0)

a < 0

H ^ m so (1) :

• D6ng bie'n t r e n khoang (- oc; 0)

• Nghieh bie'n t r e n khoang (0; +

• D 6 t h i la mot parabol c6 :

D i n h S(0 ; c)

- True doi xiJng 1^ Oy

7

a > 0 2) D a n g y = a x % bx + c (a ^ 0) (2)

B a i 1 a) Cho t^ip A c6 n phan tiif (n S: 1) Chtfng minh A c6 2" t^p con

b) Cho A = |1, 2, 3, 4, 5, 6| Gpi B la tgp con ciia A sao cho B chiiTa 1

ma khong chuTa 2 Hoi c6 bao nhieu tap B ?

• Ta CO P ( l ) diing vi neu A c6 mot phan tilr t h i A c6 2^ = 2 tap con, do

la tap 0 va A ,,

• Gia sii menh de P(n) diing k h i n = k (k > 1, k e N) ta chuTng minh P(n) diing k h i n = k + 1

- T h a t vay, k h i A c6 k p h i n tiJf, ch^ng han : A = Au = |ai, a2, , aki

t h i A CO 2^ tap con, dp la | |, |ail, |a2l, , |ai, 32, , aiJ

Do do k h i A diioc them mot phan tiir (au + i ch^ng han) t h i A c6 t h # m 2'' tap hop con bfing each them phan tuf ak + 1 vao cac tap con n6i tren, do la |au + 1), |ai, ak + i l , , l a i , a2, , ak, ak + i l

Vay so' tap B = so tap con ciia A' = 2^ = 16

B a i 2 Cho bleu thii-c T = J^J— +

• Chu y : a^ = a ^ va (x - y)^ = x^ - 2xy + y^

T xac dinh k h i mau ^ 0

• y - - 6y + 8 = 0 " y = 2 => x = 3

y = 4 i ^ y : - 3

D a p so : B(3 ; 2) va C(~ 3 ; 4)

Phiicfng t r i n h ( A B ) c6 d a n g y = k x + b, ( A B ) qua A d ; 6) v a B ( 3 ; 2) [6 = k + b

x > 0 ) D a p so': C(3; 0)

131

a) Giai giong bai 3

b) So nghiem ciia phijang t r i n h l a so giao diem ciia do t h i h a m so da

b) T r e n k h o a n g (0 ; 2) h a m so' (1) t S n g h a y g i a m

HUcfng d&n

a) Dat f(x) - T i m tap xac dinh D cua h a m so

T i n h f(-x) theo x r o i so sanh f(-x) v^ f(x)

b) Lay x i , X2 e (0 ; 2), t i n h y i = f(xi) v^ y2 = f(x2) roi t i n h y i - y2

Cho x i < X2 r o i t i m dau cua y i - y2

Ta CO X e [- 2; 21 - X e [ - 2 ; 21

• f(-x) = ^2 + (-x) + ^2 - (-x) = 72 - X + 72 + X = fix) Vay h a m so (1) l a h a m so chSn nen do t h i c6 true doi xu'ng l a true tung (dpcm)

Hai duang thing (d) : y = ax + b va (d') : y = a'x + b'

• (d) va, (d') cdt nhau o a ^a'

Vdri m * 4 t h i d i , da, ds t a o t h a n h t a m giac A B C v u o n g t a i A ( - 1; 2)

B a i 12 T r o n g m a t p h S n g toa dp, cho A( ^ 3 ; 2), B(0; 1) v a difcfng t h S n g (d) :

Cho (P) qua diem (- 4; 5) => => a =

=> T a m gi^c OBO vuong can t a i 0(0; 0)

B a i 14 Trong mSt p h i n g tpa dp Oxy, cho parabol (P) : y = ax^ + bx + c co

dinh S(2; - 1) va cdt true tung tai diem C c6 tung dp yc = 3

a) Tinh a, b, c

b) T r e n (P) lay diem A c6 hoanh dO XA = 3

Chtfng minh tam giac SAC vuong va tinh chu vi tam giac S A C

• HU&ng d&n

a) * 0(0; 3) e (P) nen • » c = 3

_b_ - b ' + 4ac' 2a

b) Tim diem I tren true tung sao cho chu vi tam giac l A C dat gia tri nho nha't

K i e n thufc c d b ^ n Cho hai diem A, B va difofng thdng (d) Tim diem M tren (d) sao cho

MA + MB ngSn nha't

A

• Trifdng hpp 1 : A va B d hai phia doi vdfi (d)

• M A + M B > AB (hang so) Vay m i n ( M A + M B ) = A B M = MQ (Mo la giao diem ciia AB va (d))

• Tri/cfng hpfp 2 : A va B of cung mpt phia doi vdri (d)

• Gpi A' la diem do'i xiJng ciia A qua (d)

( A ' C ) eat t r u e t u n g t a i d i e m Io(0 ; 13

D a p so :

I e Oy v a c h u v i t a m g i a c l A C n h o n h a t

B a i 16 T r o n g m a t p h S n g toa dp, cho ba d i e m : M ( - l ; 1), N(4; - 2 ) , P ( l ; 3)

a) T i m d i e m I t r e n true t u n g sao cho I M + I N n g a n n h a t

b) T i m d i e m J t r e n true h o a n h sao cho J M + J P nho n h a t

• Dung cong thilc I a I =

- a neu a < 0 de l a m mat dau t r i tuyet doi ciia (1)

• Khao sat h&m so' y = ax + bx + c trong moi triTdng hop

-1*. - 1 ^

+ 00

Phi/cfng trinh (^'^) c= m = - x^ + 2| x I : day la phiTOng trinh hoanh do

giao diem ciia do thi ham so y = - x % 2| x I va dtromg thfing y :r ni

24

l i a i 23 T i m t r e n do thi (P) cvia h a m so y = - + 4x h a i d i e m A v a B sao

cho A B c u n g phifofng vtJi true h o a n h v a dp d a i doan t h ^ n g A B = 6

• HUcfng d&n

• A B ciJng phuang v d i Ox nen yA = y n = m : A(X] ; m), B(X2 ; m)

• A , B G (P) nen m = - x^ + 4 x i ni = - x.^ + 4 x 9

=> X i , X2 la nghiem phuang trinh m = - x^ + 4x h a y x~ - 4x + m = 0

• A B " = ( x i - X2)'^ + (m - m)^ = ( d u n g dinh l i V i e t do'i v d i phiWng

trinh bac h a i )

G I A I

• V i A B cijng phycfng v d i O x nen y ^ = y g = m : A ( x i ; m), B(X2 ; m)

• A , B e (P) nen m = - x ^ + 4x] va m = - x^ + 4 x 2

=> x i * X 2 la nghiem phuang trinh m = - x" + 4x h a y x" - 4x + m = 0 (")

• D i e u kien de phifong trinh C*) c6 h a i nghiem phan biet X i , X2 l a

Do t h i la parabol (P) c6 dinh S{- 2; - 1), true doi xufng x = - 2, cSt

cac true toa do t a i (0; 3), (- 1; 0), (- 3; 0)

b) B (XB; YB) va S ( - 2; - 1) doi xdng nhau A ( - 3; 1) nen A la trung diem

a) Khao sat suf big'n thien va ve do thi (P) cua ham so' y = - - 4x

b) Goi S la diem cijfc dai cua (P) Tim diem M e (P) sao cho I(- 3; 2) la

trung diem doan thSng SM

ChuTng minh rSng A, B, S i , S2 la bon dinh cua mpt hinh vuong

• Hiic/ng dan

• T i m toa do Si va S2 DCing cong thufc S

T i m toa do A va B : Giai h? phUcfng t r i n h

2a 4a, [(Pi)

[(Pa)

Chu-ng m i n h tut giae A S 1 B S 2 c6 hai dudng cheo A B va S1S2 : b^ng

nhau, vuong goc nhau, eo chung mpt trung diem

B a i 28 Trong mat phang toa dp Oxy cho hai parabol ( P , ) : y =

( P 2 ) : y =

x'^ + 4x

X - 4x Chu"ng minh rang hai dinh ciia ( P i ) va ( P 2 ) cung vofi hai giao diem

ciia ( P i ) va ( P 2 ) tao thanh mpt hinh thoi

^ Hii&ng ddn ' Giai gio'ng bai 27

Dap so : S i ( - 2 ; - 1), S 2 ( - 2 ; 1), A ( - 4 ; 0), 0(0; 0)

B a i 29 Trong mat phSng toa dp Oxy, cho hai parabol :

( P , ) : y = - 2mx + + 2

( P 2 ) : y = - x ^ - 6mx - 9m^ + 2

a) Chii'ng minh rang cac dinh S i va S 2 ciia ( P i ) va ( P 2 ) cung (i trc^n

mpt duong thiing co dinh

b) Xac dinh m de S i , S 2 va 0(0; 0), A(0; 4) la bo'n dinh ciia mpt hinh

thang can day OA va S 1 S 2 nhifng khong phai la hinh chu" nhat

• HU&ng ddn

• T i m toa do S i , S2 ta se thay ket qua

• T r u n g trirc ciia doan thfing OA cat duo'ng thSng (SiS^) t a i I , de bai

thoa man k h i I la t r u n g diem ciia 8182 dong thdi 8182 ^ OA

a) Khao sat sii bien thien va ve do thi ( P ) ciia ham so y = x^ + 4x - 5

b) ( P ) cat true hoanh tai A va B (XA < 0) DUcfng thSng y = - 5 cat ( P ) tai C va D ( X D < 0) Chiing minh rang ABCD lal hinh thang can

HU&ng ddn

• T i m toa do A va B (giai phiTcfng t r i n h + 4x - 5 = 0)

• T i m toa do C va D (giai phuong t r i n h x" + 4x - 5 = - 5)

Chilng m i n h AB va CD co chung mot dUdng t r u n g trUc va A B 1^ CD

Dap so : A ( - 5; 0), B ( l ; 0), C(0; - 5), D ( - 4; - 5)

PhUo'ng t r i n h dudng t r u n g trUc chung cua AB va CD la x = ~ 2

X = - 2

0 0

C h u y e n d « 2

PhiTofng t r i n h v a bat phufoTng t r i n h b a c n h a t mpt a n

Phiidng trinh bac n h a t :

(Hoc sinh can phan biet sir khac

phuong t r i n h bac nhat)

Bat phadng trinh bac nhat ;

( c h u y : chia hai ve cho so' diicfng

t h i bat philcfng t r i n h khong doi chieu, chia hai ve cho so am t h i bat phucfng t r i n h doi chieu)

* A = 0 liic do bat phi/cfng t r i n h tror

t h a n h 0.x > B, 0.x > B, 0.x < B, 0.x < B

- Tuy theo dau cua B va dau bat dang thiJc ta duac tap nghiem ciia bat phucfng t r i n h la S = R hoac

S = 0

M e n h de Chan t r i 0.x >, < 0 Sai 0.x > < 0 Dung 0.x >, > so dacfng Sai 0.x <, < so diicfng Diing 0.x >, > so am Dung 0.x <, < so' am Sai

Giai giong bai 31 ta c6 : n i ( m + 2 ) x = ( m + 2 ) ( m - 2 )

m Tap nghiem ciia phiicfng t r i n h

<=> x^ + 2 m x + m^ + 1 0 m + 6 = 5 ( m ^ + 2 m + 1) + x^ - x

o ( 2 m + l ) x = 4 m ^ - 1 = ( 2 m + l ) ( 2 m - 1) Phifcfng t r i n h c6 t a p n g h i e m S ^ R v a S # 0 ( n g h i a l a phUcfng t r i n h

B a i 36 Giai v a b i ^ n l u ^ n theo m b a t phi^oTng t r i n h :

• 2 - m > 0 o m < 2 ; bat phacfng t r i n h c6 nghiem x > - (m + 2)

• 2 - m < 0 o m > 2 : bat phaang t r i n h c6 nghiem x < - (m + 2)

• 2 - m = 0 o m = 2 : ba't phaong t r i n h C'') t r d thanh 0.x > 0 : v6 nghiem

B a i 37 Tuy theo a, giai va bi§n lugin bat phifcfng trinh :

2(a - 1) + (x + 1)^ < a^d - x) + 3(a - 1 + X) + x^ (1)

CO. (a - l)(a + l)x ^ (a - l)(a + 2) (*)

• Trifdng h^p 1: a^ - 1 > 0 o I a I > 1, luc d6 bat phacfng t r i n h (*)

= be' - b'c, Dy = c a = ca' - c'a

• T r i f d n g hdp I : D Q c : > n i ; t ± l , he phiTOng t r i n h c6 n g h i e m

m duy n h a t

X =

D D„

• Trircfng hop 1 : D 0 (khong xet D^, Dy)

• Trifdng hcfp 2 : D = 0 (xet D^, Dy)

a'x + b'y + c' = 0 CO nghiem

(qi - q2)^ + (pi - P2)(piq2 - P2qi) = o (*)

Can nhd: He phUang trinh

ax + by + c = 0 a'x + b'y + c' = 0

C o n g (4), (5), v a (6), t a c6 : a^ + b^ + c^ = = ab( ) + bc( ) + ca( )

B a i 50 Chu'ng minh h$ phi^cfng trinh : a x - y - 2 a + l = 0

3x +(a - 2)y + a - 3 = 0 luon CO nghi$m

X X

a b 3) a.x < b.x ; — < — neu x < 0

C^c phiromg pbAp gial bai loan bat dAng thurc

• SiJf dung cac kien thiJc cd ban de chuTng m i n h mot bat dSng thufc

• ' Suf dung bat d^ng thiJc Cosi ,

• iTng dung dau " = " xay ra de t i m gia t r i Idn nhat, gia t r i nho n h a t

ciia mot bieu thiic

• a, b la hai so tuy y, ta c6 : a' + b > 2ab (1)

• PhUdng phap tacfng ducfng

Muo'n chijfng m i n h bat d^ng thufc (1) dung ta c6 the t i e n h a n h n h u sau :

BrfdTc 1 : T i m bat d^ng thiJc tirang difang

Bat dang thufc (1) o o o bat d i n g thilc (2)

BrfdTc 2 : Chufng m i n h bat d^ng thiic (2) dung, tii do ket luan bat

d i n g thufc (1) diing

Chii y :

Bat d i n g thufc (2) hoSc d i thay dung, hoSc la bat d i n g thufc thudng

gap, hoac dung nh6 phu hcfp vdi gia thie't

B a i 51 Chufng minh rfing v6i nam so' a, b, c, d, e bat ki, bao gid ta cung

'=> Bat d i n g thufc (2) dung

=> Bat d i n g thiJc (1) dung (dpcm)

y - z < 0

B a i 53 Cho a, b, c la cac dp dai cac canh ciia mpt tam giac

Chufng minh rkng : a^ + b^ + c^ < 2(ab + be + ca)

^ Hitdng dan

a, b , e > 0

• a, b, c la ba canh ciia mot t a m giac nen : < a < b + c => a^ < ab + ca

Cong ve theo ve cac d i n g thiifc t i m duac

45

B a i 54 Chiifng minh rSng neu a i a 2 > 2(bi + b^) thi it nhat mpt trong hai

phUcfng trinh sau c6 nghipm :

+ aix + bi = 0 (1)

x'* + aax + ba = 0 (2)

• HUdng ddn

• T i n h biet so' Ai va A2 ciia hai phuang t r i n h

• Chufng m i n h Ai + A2 > 0 bkng each suT dung bat 'd^ng thufc

a^ + b'^ > 2ab roi suf dung gia thiet

Chii y : a + b > 0 => Trong hai so a, b c6 i t nhat mot so > 0

Trong so Aj va A2 c6 i t nhat mot so' khong am

=> Trong hai phiTcfng t r i n h c6 i t nhat mot phucfng t r i n h c6 nghiem

B a i 55 Chuing minh r^ng neu : I x I < 1, n la so nguyeii va n > 1 thi ta c6

bat dang thuTc : (1 - x)" + (1 + x)" < 2"

* HU&ng ddn

• 0 < a < 1 va n, p G N n > p a" < a"^

1 - X 1 + X

Biet so' cua phiicrng t r i n h

Tii gia thiet, chufng m i n h 0 < 1

Cong (1) va (2) ve theo ve ta c6 :

1 - x l 1 + X

< 1 o (1 - X ) " + (1 + X ) " < 2" (dpcm) •

B a i 56 a, b, c la dp dai ba canh ciia mpt tam giac

Chufng minh rdng : abc > (a + b - c)(a + c - a)(a + c - b)

• Chufng m i n h xy + yz + zx > - ^ b^ng each chii y :

Ve phai = - - + y^ + z'^), sau do bien doi bat dang thufc v l dang

2 (x + y + z)^ > 0

47

o (2 + a'^ + b^)(l + ab) > 2(1 + a^)(l + b^) (vi ab > 1 nen 1 + ab > 0)

o 2 + a^ + b'^ + 2ab + a^b + ab^ > 2 + 2a^ + 2b^ + 2a^b^

o ab(a^ + b^ - 2ab) - (a^ + b^ - 2ab) > 0 « ab(a - b)^ - (a - b)^ > 0

Nhinig 1 + V ^ l + Jab^ 1 + ^a'^b'^abc" 1 + abc 1 1 + 2

(theo ket qua phan 1)

=> B a t d i n g thufc (4) diing => Bat d i n g thiJc (3) diing (dpcm)

B a i 61 Cho 0 < a, b, c < 1 Chtfng minh rSng c6 it nhat mpt trong ba bat

d^ng thiic sau day sai :

64 (1)

• Gia siif ca ba bat d i n g thiic neu t r e n deu diing, tii do suy ra mot bat

d i n g thufc mau t h u i n vdi bat d i n g thiJc (1)

Vay CO i t nhat mot trong ba bat dSng thufc da cho la sai

B a i 62 Cho a, b, c e [- 1; 2] thoa man dieu ki^n a + b + c = 0

<=> x^ + y^ + z^ < 5 - xyz < 5 (vi xyz > 0)

B a i 64 Cho a, b, c la ba so tuy y Chiifng minh :

S a i 65 a, b, c la dp dai ba canh ciia mpt tam giac va p la nijfa chu vi ciia

tam giac do ChuTng minh rdng :

abc a) (p - a)(p - b)(p - c) <

8 b) 1 - i - - l - 2

b) TrUdc het, chufng m i n h rang vdi x > 0, y > 0 ta c6 :

N h a n ve theo ve cac bat dang thtrc (1), (2) va (3) ta dUdc :

p - a)" (p - hf (p - c)^ < z=> Bat d i n g thdc (i) dung (dpcm)

, Phan tich 3a^ + 7b^ = 3a^ + 3b^ + 4b^ r o i ap dung bat d i n g thufc

Cosi cho ba so khong a m 3a^ 3b^ 4b^ s

• Tir gia thi§'t, t i n h b theo a va c ,

• T i n h ve t r a i ciia bat dang thufc da cho theo a va c (thay b v t o t i n h vao ve t r a i ciia bat dang thufc da cho)

• SiJf dung bat dSng thufc Cosi doi vdi hai so diTcfng va

-55

• Lan iMt ap dung :

Bat d4ng thdc Cosi cho ba so' difcfng :

B a i 69 C h o tam giac A B C c6 b a goc n h o n G p i H la trtfc tanj tam giac,

c h a n c a c dtTorng cao ke tU A, B , C l a n li^ort la A j , B j , C i

- A H B H C H ^ „ Chu-ng mxnh r a n g : ^ + ^ + ^ 6

K h i n a o thi da'u d S n g thiiTc x a y r a ?

1^ Hiidng ddn

• Bvtdc 1 : T i m bat dSng thiJc tUOng dircfng b^ng each liAi y : '

• H d m i l n trong tam giac ABC nen : A H = A A i - A i H

A H ^ A A i - A i H ^ A A ^ _ ^ ^ S,,ABC _ ^

A j H A j H A j H 5,\HBC Bien doi bat dSng thiJc da cho ve dang ;

B j H S, (Chu y S = Si + S2 + Sg) Vay bat dang thurc da cho

Dau "=" xay ra o S i = S2 = S3 o Tarn gidc ABC la tam giac deu

Bai 70 Cho a, b, c> 0 Chiifng minh :

b + c > 4a(b + c)^ > 4a(2N/bcj^ => b + c > 16abc (dpcm)

Bai 72 Cho x, y > 0 va x + y = 1 Chtfng minh

59

Cho hai so' x, y thay doi va thoa man dieu ki$n 0 < x < 3 v a 0 < y < 4

Tim gia tri idrn nhat cua bieu thuTc A = (3 - x)(4 - y)(2x + 3y)

• HUdng dan

• Viet l a i : A = ^ (6 - 2x) - (12 - 3y)(2x + 3y)

(Muc dich (3 - 2x) + (12 - 3y) + (2x + 3y) = hang so)

• Ap dung bat dang thuTc Cosi cho ba so' khong am, ta se c6 A < hSng so'

• Phan tich f t h a n h dang tong ciia ba so' A, B, C

• Sii' dung bat dang thiJc Cosi Ai chuTng minh A < hfing so', tiT do

tinh dUOc

fmax-G I A I

'a - 3 7 ^ - 4 Viet l a i

Ta CO (theo bat dang thilc Cosi : 4ah < a + b

* T i n h y^ va ap dung bat d i n g thiJc Cosi 2 Tab < a + b ta se chdng

m i n h dugc y^ < hang so (chii y : y > 0)

b) PhLTOng trinh yjx - 2 + ^4 - x = x^ - 6x + 11 (*)

• Ve trai = - 2 + ^4 - x < 2 (DS'u = xay ra o x = 3)

• Ve phai = x^ - 6x + 11 = (x - 3)^ + 2 > 2 (Dau = xay ra o x = 3)

f(x) = 0<=>ax + b = 0 <P> X =

a c) Dau :

ax -f b (trai dau v d f i a) 0 (cung dau v c r i a)

Cdch nhd : Phai cung, trai trai

• Phufdng phap giai toan Xet dau n h i thufc bac nhat f(x) = ax + b

• T i m nghiem (giai phiTcrng t r i n h ax + b = 0) Lap bang xet dau

a + 00

ax + b trai dau vdi a 0 cung dau vdi a

• Ghi ket qua

B a i 79 Xet dau nhi thu-c a) f(x) = 2x - 3 b) f(x) = - 4 - 3x a) f(x) = 2x - 3 (a = 2 > 0)

c) f(x) = 1 - - X

2 d) f(x) = 2x + l GIAI

• Nghiem : 2 x 3 = 0 < z > x =

-63

• Bang xet dau :

2 f(x) = - 4 3 x (a = - 3 < 0)

N g h i e m : - 4 3x = 0 <:> X : 4

3 Bang xet dau

2 2 Giai van t d t

c h u y e n 6 PNcTONG T R I N M B ^ C net MpT en

Kien thufc cd ban

Phuang t r i n h bac hai : ax^ + bx + c = 0 ( a ; * 0 )

• Biet so' A = b^ - 4ac hoac A' = b'^ - ac (b' = — )

• a + b + c = O o Phifcfng t r i n h c6 hai nghiem

• a - b + c:=0 Phifdng t r i n h c6 hai nghiem

Gia sijf phuong t r i n h ax^ + bx + c = 0 c6 hai nghiem x i , X2

• Tong hai nghiem S = X i + X2 = - —

(Hoc sinh tif ghi cac ket qua con lai)

ChuTng minh phUdng trinh b|ic hai iuon c6 hai nghiem

Vay phuang t r i n h luon c6 hai nghiem phan biet • X j = - m + 1

• X g = - m - 1

Tfnh gia trj cua tham so de phUdng trinh c6 nghiem l<ep va ti'nh

nghi?m kep ti/dng ufng

* Hudng ddn

• T i n h A ho&c A' (theo tham so)

• Phuong t r i n h c6 n g h i f m kep o A = 0 (hoac A' = 0)

Nghiem kep ciia phuong t r i n h la : X i = X2 = — ^ hoac x i = X 2 =

69

Tinh nghiem con l a i

Xac dmh dieu kien cua mpt tham so de phUdng trinh bac hai c6

hai nghiem hoac vd nghl#m

B a i 91 Cho phvtcfng trinh x'^ - 2(a + b)x + a'' + = 0

Tim s\i lien h$ giffa a va b de phvtdng trinh c6 hai nghi^m phan bi$t

• HU&ng ddn

A' = 2ab

A' > 0 <=> ab > 0 <=> a va b cung dau

Tinh gia tn c u a mpt bieu thufc doi xtifng doi vdi c a c nghiem x i , X2

c u a phi/dng trinh b a c hai : ax^ + bx + c = 0 {a^O)

E = x f + , E = X ? + , E = — + — ,

X i X2

B a i 92 Gia si3f phxfofng trinh x^ - (m - l)x + 2m + 1 = 0 c6 hai nghi^m x i , x^

Tinh cac bieu thufc sau theo m

• T i n h S va P roi thay vao bieu thufc E

• Chu y : a^ + b^ - (a + b)(a^ + b^ - ab) = (a + b)[(a + b)^ - Sab]

Chti y : Bai nay de khong yeu c l u xac dinh m de phiTofng t r i n h c6 hai

B a i 93 Cho phvLdng trinh (m + Dx" - 2(in - l)x + m - 2 = 0 (*)

a) Xac dinh m de phUdng trinh (*) c6 hai nghiem

b) Vdri nhu'ng gia tri nao ciia m thi phufcfng trinh (*) c6 hai nghi$m

xi, X2 thoa man : 4(xi + X2) = 7 x i X 2

b) Tim h $ thxJc giffa X i , X2 dQC l $ p doi vdi m

B a i 95 Cho phUoTng trinh (m - l)x^ + 2(m + 2)x + m - 1 = 0 Tim m de :

a) Fhi/oTng trinh c6 dung mpt nghiem

b) Phtfcfng trinh c6 hai nghiem cung dau

* Hudng ddn

a) PhiTcfng t r i n h c6 dung rapt nghiem

r Phacfng t r i n h t r d t h a n h phiforng t r i n h bac nhat

B a i 96 Cho phUcfng trinh mx^ - 2(m - 3)x + m - 4 = 0 (*)

Xac dinh m de phifcfng trinh c6 dung mpt nghiem dUofng

• Hiidng ddn

Phuang t r i n h (*) c6 dung mot nghiem dircfng

Phircfng t r i n h (*) la phiicfng t r i n h bac n h a t c6 nghiem difomg (1)

Phiicfng t r i n h (*) c6 h a i nghiem t r a i dau (2) hoac c6 mot n g h i e m

va nghiem con l a i diTcfng (2')

- Phuang t r i n h (*) c6 nghiem kep duong (3)

Phucfng t r i n h (*) c6 mot nghiem x = 0 o m - 4 = 0 c : > m = 4 1

fx = 0 , Vdi m = 4 t h i phucfng t r i n h (*) la 4x^ + 2x^ = 0 o _ 1

~ '2

Vay m e n h de (2') k h o n g xay ra

9 •

Dap so : De bai thoa man < : = > 0 < m < 4 V m =

-Can nh& : Phuang trinh bac hai ax' + bx + c = 0 (a ^0) c6 hai nghiem

trai dau x i < 0 < xo P < 0 (hie do hien nhien A > 0 vi P = — < 0

a

B a i 97 Cho phtfc*ng trinh x^ - 2(m - 2)x + m(m - 3) = 0 (1) a) Xac dinh m de phifofng trinh (1) c6 hai nghiem phan bi$t X i , X 2 thoa

-b) Chiing minh rSng neu phifdng trinh tren c6 mpt nghiem dtfoTng la

xi thi phUoTng trinh : m(m - 3)x^ - 2(m - 2)x + 1 = 0 cung c6 mpt

1

n g h i $ m diicfng la —

X ,

75

B a i 98 Cho phUorng trinh 2x^ - 2(m + l)x + m + 5 = 0 (1)

Xac dinh m sao cho phvTdng trinh (1) c6 hai nghiem xi, X 2 thoa man

B a i 99 Cho phirong trinh 3x^ - (2m + l)x - 2 = 0 (*)

a) Chiang minh phi^ofng trinh (*) luon c6 hai nghiem phan bi§t b) L 4 P phifofng trinh b^c hai c6 hai nghi$m ai, a2 thoa mSn aj

B a i 100 Cho hai phifcfng trinh : x^ + mx + l = O v a x ^ + x + m = 0

Xac dinh m de hai phifofng trinh c6 mpt nghiem chung

x^ + y^ = (X + yXx^ - xy + y^) = (x + y)[(x + y)^ - 3xy|

i i 103 Giai h § phtfoTng t r i n h : X + y = 5

X * + y^ = 97 GIAI

xUy' = ix' + y^)^ - 2 x V '

Chii y : (xy) - 50(xy) + 264 = 0 l a mot phiicrng t r i n h bac hai a n so l a (xy)

Ta CO hai truomg hop :