數學4題(恆等式)

(3x-2)^2-3≡ax^2+bx+c,求ａ，ｂ，ｃ

(3x-2)^2-3
=9x^2-12x+4-3
=9x^2-12x+1
so, a=9, b=-12, c=1

a(x+2)(x-2)+b≡2x^2,求a和b

a(x+2)(x-2)+b
=a(x^2-4)+b
=ax^2-4a+b
so, a=2 and -4(2)+b=0,b=8
so,a+b=2+8=10

2007-01-15 18:49:09 補充：
(3) no solution------------------------------------------------------------------------------------------a(x-2)(x+2)+b≡-x^2,求a和b的值a(x-2)(x+2)+b=-x^2a(x^2-4)+b=-x^2ax^2-4a+b=-x^2so,a=-1,-4(-1)+b=0,b=-4 so a=-1,b=-4

(1)若(3x-2)^2-3≡ax^2+bx+c,求ａ，ｂ，ｃ
LHS:(3x-2)^2-3
=9x^2-12x+4-3
=9x^2-12x+1
comparing the coefficient:
a=9, b=-12, c=1

(2)若a(x+2)(x-2)+b≡2x^2,求a和b
LHS:a(x+2)(x-2)+b
=a(x^2-2^2)+b
=ax^2-a2^2+b
comparing the coefficient,
a=2,
-a2^2+b=0
-2(4)+b=0
b=8
thus, a=2, b=8

(3)若a(x-1)^2+b(x+2)≡3x^2+2,求a和b
LHS:a(x^2-2x+1)+bx+2b
=ax^2 - 2ax +a +bx +2b
=ax^2 +(b-2a)x +(a+2b)
comparing the coefficient,
a=3,
b-2a=0..............(1)
a+2b=2................(2)
2*(1)-(2),
2b-4a-(a+2b)=0-2
-5a=-2
a=2/5
sub. a=2/5 into (1)
2/5+2b=2
2b=-8/10
b=-2/5
thus,
我計5到~5g why有2個答案= =(係咪問題出錯?.?)

(4)若a(x-2)(x+2)+b≡-x^2,求a和b的值
LHS:a(x+2)(x-2)+b
=a(x^2-2^2)+b
=ax^2-a2^2+b
comparing the coefficient,
a=-1
-a2^2+b=0
-(-1)4+b=0
b=-4
thus,
a=-1,b=-4

(1)
(3x-2)^2-3≡ax^2+bx+c
9x^2-12x+4-3≡ax^2+bx+c
9x^2-12x+1≡ax^2+bx+c
所以a=9,b= -12,c=1
(2)
a(x+2)(x-2)+b≡2x^2
a(x^2-4)+b≡2x^2
ax^2-4a+b≡2x^2
所以a=2,
-4(2)+b=0
b=8
(4)
a(x-2)(x+2)+b≡-x^2
a(x^2-4)+b≡-x^2
ax^2-4a+b≡-x^2

所以a=-1,-4(-1)+b=0,b=-4