a) y = 7 x ^ - 2 x + 3 1
X - 3Vx + 2
f x 2 - 2 x + 3>0 a) DKXD: \
[ x - 3 V x + 2 ^ 0 Suy ra D = 1R\{1;4}
b) f(x) =
V i - V T + ^ Huang dan gidi
( x - l ) ^ + 2 > 0 ( V ^ - l ) ( 7 ^ - 2 ) ^ 0
x ^ l
X ^ 4 Or
b) DKXD: • 1 - >yi + 4x > 0 l + 4 x > 0
l > l + 4x
1 <=> — <x<0=>D = 1 x > — 4
4
51
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 a) D K X D : x m
H a m so xac dinh tren (-1;0) <=> m ô ( - 1 ; 0 ) <=> m > 0 m < - 1 b) D K X D : I ^ ^ ^ (*)
[x > m
Neu m > 0 thi n ô 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 -
a) m > 2 , b) m e [ 0 ; ] ] Vx + m
Huang dan gidi
c) m 6[ l; 3 ] DANG TOAN 2: XET TINH CHAN, LE CIJA HAM SO
• H a m so chSn
• H a m so'le <=> • 'Phuxmg phdp gidi.
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.
C a i. C A C V I D U M I N H H Q A ' ^
V i du 1: Xet tinh chan, le ciia cac ham sosau:
a) f(x) = 3x^ + 2 ^ b) f(x) = x ^ + 7 x^ + 1
Led gidi a) Ta C O TXD: D = M
Voi moi X € K ta c6 - x € IR va f ( - x ) = 3 ( - x f + 2 ^ = -(3x^ + 2 ^ ) = - f ( x )
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) Do do f(x) = x"* + Vx^ + 1 la ham so chan
V i du 2: Xet tinh chan, le cua cac ham so sau:
b) f(x) = a) f(x) = ^ i l £ ± l - 2 x ^ - 1
- 1 K h i X < 0*"
0 K h i X = 0 1 K h i X > 0 Led gidi
a) Ta C O Vx^ + 1 > Vx^ = |x| > x => N/X^TT - X 0 voi moi x.
Suy ra T X D : D = M
Mat khac V x ^ T l > Vx^ = Ixl > - x Vx^+ \+x^0 do do
[f(x) =
X +
X + 1 + X V ? T i - x )
-1x^-1 = 2x4x^+1
I V o i m o i X e M ta c6 - x e M va f ( - x ) = l{-x)^{-xf+1 = -2xVx^ +1 = - f (x)
Phdtt loai vd phuong phdp gidi Dai so 10
D o do f(x) = y _ _ ,^ j ^ ^ j ^ 1^
Vx^ + 1 - X
b) T a c o T X D : D = IR
De thay m o i x e M ta c6 - x e R - >.
V 6 i m p i x > 0 ta c6 - x < 0 suy ra f ( - x ) = - l , f (x) = 1 => f ( - x ) = - f ( x ) V a i m o i x < 0 ta c6 - x > 0 suy ra f (-x) = l , f ( x ) = - 1 => f ( - x ) = - f ( x ) Va f ( - 0 ) = - f ( 0 ) = 0 ,
D o d o v o i m o i x e M ta c6 f ( - x ) = - f ( x ) f - 1 K h i x < 0
0 K h i x = 0 la ham sole.
1 K h i X > 0 Vay h a m so f(x) =
x2(x2- 2 ) + ( 2 m 2 - 2 ) x
V i d u 3: T i m m de h a m so: f ( x ) = —!^ = L = A ^ h a m so c h i n .
Vx^ + l - m Lai gidi
D K X D : 7 x 2 + 1 ^
Gia s u h a m so chan suy ra f (-x) = f (x) v a i m p i x thoa m a n dieu kien (*) x 2 ( x 2 - 2 ) - ( 2 m 2 - 2 ) x
Ta C O f (-x) = —5^ —l—^ '—
V x ^ + l - m
Suy ra f ( - x ) = f (x) v a i m p i x thoa man dieu kien (*) x 2 ( x 2 - 2 ) - ( 2 m 2 - 2 ) x x^ ( x ^ - 2 ) + ( 2 m 2 - 2 ) x
/ = , v a i m p i x thoa man V x ' ^ + l - m V x ^ + l - m
dieu kien (*)
o 2^2m2 - 2Jx = 0 v a i m p i x thoa m a n dieu k i ^ n (*)
<=>2m2 - 2 = 0 m = ±1
V a i m = 1 ta c6 ham so la f (x) = ^ ' V x ^ + l - l D K X D : \ / ) ? T l ^ l o x ^ O
Suy r a T X D : D = IR\{0}
De thay v a i m p i x € M \} ta c6 - x e K \} va f (-x) = f (x) D o d o f ( x ) = ; ' la ham so c h i n
V x ^ + l - l
x^ ( x 2- 2 )
* V a i m = - 1 ta c6 h a m so la f (x) = - j = =
TXD: D = IR
DI thay v o i m p i x e M ta c6 - x € M va f (-x) = f ( x ) D o d o f ( x ) = - 7 = = - ^ la h a m so chan.
N/X^ + 1 + 1 V^y m = ±1 la gia t r i can t i m .
p 2. BAI TAP LUYEN TAP
Bai 2.4:: Xet tinh c h i n , le cua cac h a m so sau:
a) f(x) = > / ^ - V T ^ b ) f ( x ) =
c) f(x) = 3x2- 2 x 4 - 1 d ) f ( x ) = x - 5 x - 1 x^
N - 1
. mi. •
| x - l | + |x + l | „ . |x + 2| + |x-2|
" ' W' | 2 x - 1 M 2 x . l | " ' W ' F F M
Huang dan giai a) TXD: D = [-!;!] nen Vx € D - x G D
f (-x) = - ^/TTx = - f ( x ) , Vx G D Vay h a m so da cho la ham so le.
b) TXD: D = R\{1}
Ta CO x = - 1 e D n h u n g - x = 1 g D Do d o h a m so khong chan va khong le c) TXD: D = K . Ta C O f (1) = 2, f (-1) = 6
Suy ra f (-1) ^ f (1), f (-1) ^ - f (1) Do d o h a m so khong chan va khong le.
d) T X D : D = ( - o o - l ) u ( - l ; l ) u ( l ; + o o ) nen V x € D = > - x 6 D f ( - X ) = = = - f ( x ) , Vx 6 D
- x | - l |x - 1
Vay ham so da cho la ham so le.
e) T X D : D = M = > V x G D = > - x e D .
r/ ^ | - X - 1 | + | - X + 1| w f ( - x ) = -! L J L = f ( x ) , Vx G D
.3
- 2 x - l | + |-2x + l |
Vay h a m so da cho la h a m so chan.
f) DKXD: |x-l|9^|x + l|<=> X - 1 ?t X + 1 x - 1 ^ + TXD: D = K \} ^ Vx e D => - X e D
- X + 2 + |-x - 2
f(-x) = = - f ( x ) , V x e D
| - x - l | - | - 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 va f (-x) = - f (x), g(-x) = -g(x)
suy ra y ( - x ) = f (-x) + g(-x) = - [ f ( x ) + g(x)] = - y ( x ) Suy ra ham so y = f (x) + g(x) la ham so le.
b) Giasuhamso y = f(x) chan, y = g(x) le
Khi do ham so y = f (x)g(x) co tap xac djnh la D nen Vx e D => -x e D Ta CO y( - X ) = f (-x)g(-x) = f (x)[-g(x)] = - f (x)g(x) = - y ( x )
Do do ham so y = f (x)g(x) le.
Bai 2.6: a) Tim m de do thj ham so sau nhan goc toa dp O lam tam doi xung y = x ^ - ( m ^ - 9 ) x 2 + ( m + 3)x + m- 3 .
b) Tim m de do thj ham so sau nhan true tung lam true doi xung y = x' ' - ( m 2 - 3 m + 2 ) x 3+ m 2- l .
Huang dan gidi a) Ta CO TXD: D = R = > V x e D = > - x e D
Do thi ham so'da cho nhan goc toa dp O lam tam doi xung khi va chi khi no la ham so' le
o f ( - x ) = - f ( x ) , V x £ R o ( - x ) ^ - ( m ^ -9 )(-x)^ + (m + 3 )(-x) + m - 3 x ^ - ( m ^ -9)x^ + (m + 3)x + m - 3 , Vxe:
o 2 ( m ^ -9)x^ - 2 ( m - 3 ) = 0, Vx€Rc:><!
b) Ta CO TXD: D = M = > V x e D = > - x e D
m ^ - 9 = 0
m - 3 = 0 <:=>m = 3
£)5 thi ham so da cho nhan true tung lam true doi xung khi va chi khi no la ham so chan
^ f ( _ x ) = f ( x ) , V x e l
^ ( _ x f - (m^ - 3m + 2 ) ( - x f + m^ - 1 = x^ - (m^ - 3m + 2)x-^ + m^ - 1 , Vx e R
^ 2(m^ - 3m + 2)x^^ = 0, Vx E R ô m^ - 3m + 2 = 0 o m = 1 m = 2
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 (!;+ô) a) y =
x - 1 b ) y . x . -
Lai Gidi a) Voi mpi X j , X2 e (1; + ô ) , x^ ; t X2 ta c6
3 3 3( x i- X 2 )
f ( x 2 ) - f ( x i ) = Suy ra
X 2- 1 X i- 1 ( X 2- I ) ( x i - 1 ) f( x 2) - f ( x i ) 3
X 2 - X 1 - ( X 2 - I ) ( x i - 1 )
Vi xi >1,X2 >]=>ii^^2HN)<o nen ham so y = - ^
X 2 - X 1 x- 1
khoang (1;+00).
b) Voi mpi X i, X 2 e (!;+<ằ), x i ; t X2 ta CO
f
nghjch bien tren
f(x2)-f{x0 = X2 + = ( X 2- X l ) X l X 2 y 1
Phan loai va phttcmg phdp giai Dai so 10
X 2 - X 1 X1X2 f( x 2) - f ( x i )
Cfy TNHHMTV in\'l! Khang Vi?t
V i X ] > 1, X2 > 1 khoang ( l; + o o ) .
X2 - X i > 0 nen ham so y = x + — dong bien tren
V i d u 2: Cho ham so y = x^ - 4
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]
X - 1 0 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.
[-•;3]' [-l;3]^
V i d u 3: Xet su bien thien ciia ham so y = V4x + 5 + V x - 1 tren tap xac djnh ciia no.
A p d u n g giai p h u o n g trinh
a) V4x + 5 + = 3 b) V4x + 5 + V x ^ = V4x^ + 9 + x Lai Giai
D K X D : 4 x + 5 > 0
x - l > 0 4 <=> x > l x > l
S u y r a T X D : D = [ ! ; + * )
V o i m o i X i , X 2 e[ l; + o o ) , xj ?t X2 ta c6
f( x 2) - f ( ' ^ i ) = V'*'<2 + 5 + N/X2 - 1 - ^ 4X1 + 5 - ^ X1 - 1 4( X 2 - X i ) ^ X 2 - X 1
^4x2 + 5 + ,y4xi +5 ^ x j - 1 + 7x1 - 1
4 1
ra
= ( X 2 - x i )
f ( x 2 ) - i ( x 0 _
^^4x2 + 5 + ^4x1 +5 ^^^2 - ^ + 7^1 ~ ^
4 1
X2 74x2+5 + 74x1+5 7^2 -1 + 7^1-1 > 0 fvjen ham so y = 74x + 5 + Vx - 1 dong bien tren khoang [ l; + c o ) . 3) V i ham so'da cho dong bien tren [l;+oo) nen
jsjg'u X > 1 => f ( x ) > f (1) hay 74x + 5 + 7 x- l > 3
Suy ra p h u o n g trinh 74x + 5 + V x - l = 3 v6 nghiem ^.
Ne'u x < l = > f ( x ) < f ( l ) hay V4x + 5 + 7 x- l < 3 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 ^ ô 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) b) y =
c) y = x - 2
X
tren ( - ô 3; 2 ) va tren (2;+GO) tren
Huang dan gidi a) Voi moi x i , X2 e x^ X2 ta c6
f( x 2) - f ( x , ) (x2 +4x2 - 5) - ( x f + 4x, - 5
K = = xj + X2 + 4
X 2 - X | X 2 - X |
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 X j,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) c) V o i moi X j ^ X j G (-=c;l), x, X2 ta c6
f ( x 2 ) - f ( x i ) =
Suy ra
X 2 X, - X 2
X 2 - 1 X i- 1 ( X 2 - I ) ( x i - 1 ) f ( x 2 ) - f ( X i ) _ - I
X 2 - X , { X 2- I ) ( x , - ] )
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
• Voi moi Xj,X2 e K, X] ?t X2 ta c6
f ( x 2 ) - f ( x i ) ( x 2 + X 2 ) - ( x ? + x 7 9
= X2 + X| + X2X] +1 > 0 X 2 - X , X 2 - X ,
Suy ra ham so da cho dong bien tren R
• Ta C O x-' - X = ^2x + 1 +1 o x'"* + x = 2x +1 + ^2x + l 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 [!;+ô:)
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.
b) * Ta C O f (x) = 1 ô 2x^ + 3x + 1 = 1 <^ 2x^ + 3x = 0 <=>
* V a i X > 2 ta C O g ( x ) = 1 ô
V o i - 2 < x < 2 ta C O g ( x )
x = 0
^ 3 2 x > 2 (x>2 x 2+ l = l l x = 0 ^ •
f - 2 < x < 2 . 2 x - l =1
- 2 < x < 2
x = l ^ x = l Voi X < -2 ta C O g ( x ) = 1 ô
Vay g ( x ) = 1 <r> x = 1 .
x< - 2 x < - 2 o < v6 nehiem 6 x - 5 = l x = l ^ •
V i dv 2: Cho ham so y = mx^ - 2(m^ + l)x^ + 2m^ - m 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
X - X +1
V i dv 3: C h u n g m i n h rang tren do thj (C) ciia ham so y = — ton tcii x + 1 h a i d i e m A ( x A ; y A ) va B ( x B ; y B ) thoa man: < 2 x A+ y A = 3
2X B + y B - 3 • Lai gidi
2 2 T a c o A e ( C ) ô y A = ^ ^ ^ ^ ^ , B 6 ( C ) ô y B = ^ ^ ^ ^ ^
[ 2X A + y A = 3 D - * * " | 2 x B . y B = 3 ^
Voi " A " B ' ' - I t3 C O
2 X A + X A - X A + 1 _
X A + 1 = 3
2 X R + X B - X B + ^ _
3 x i - 2X A - 2 = 0 3X^ - 2X B- 2 = 0 <=> <
X A = X B + 1
1±N/7 3 1± 7 7
= 3
(thoa man)
i (d)^ -ằ• • >' ••
I f f , - iS^':
Suy ra ton tai hai diem A ( x A ; y A ) va B(xB;yB) thupc do t h i (C) thoa man:
[ 2X A + y A = 3 2xB + y B= 3 •
Vi du 4: T i m tren do thj ham so y = - x ' ' + x^ + 3x - 4 hai d i e m doi x u n g nhau qua goc toa dp.
Lai gidi
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
2x^ - 8 = 0
y o = - 2 hoac
xo = ±2
XQ = - 2 y o = 2
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
3) f(0) = 5<=>m + l = 5 ô m = 4
^) Do thi cua ham soy = f ( x ) d i qua diem A ( 1 ; 0 ) khi va chi k h i m = l
0 = -3 + m ^ + m + l o m ^ + m - 2 = 0<=>
m = - 2