F.2 MATH 求恆等式中常數A & B的值

(8A+3B)x +2B=7x-6
Since x = unknown,

let x = 0,

(8A+3B)(0) + 2B = 7(0)-6, (8A+3B)(0) = 0; 7(0) = 0
2B = -6
B = -3

let x = 1

(8A+3B)(1) + 2B =7(1)-6
8A+3(-3) +2(-3) = 1
8A -9-6 = 1
8A = 1+6+9 = 16
A = 2


2(x^2-1)+Ax=(x+B)(Cx-1)

let x = 0

2(0^2-1) + A(0) = (0+B)(C(0)-1)
2(-1) + 0 = B (-1)
-2 = -B, B = 2
let x = 1

2(1^2-1)+A(1)=(1+2)(C-1)
2(0) + A = 3 (C-1)
A = 3C - 3

let x = 2

2(2^2-1)+2A=(2+B)(2C-1)
6 + 2(3C-3) = 4(2C-1)
6+6C-6 = 8C- 4
6C = 8C - 4
4 = 2C
C = 2

when C = 2, A = 3C - 3 = 3(2) - 3
A = 3

Check:
2(4-1) + (3)(2) = 4(4-1) = 12

No.1
(8A+3B)x+2B=7x-6

L.H.S.=(8A+3B)x+2B
=8Ax+3Bx+2B

so, 8Ax+3Bx+2B=7x-6

By Comparing like terms,
B=-6/2
=-3

A=(7x-3Bx)/8
=[7x-3*(-3)x]/8
=[7x-(-9)x]/8
=(7x+9x)/8
=16x/8
=2x

2006-10-28 18:19:51 補充：
NO.1Ax=(7x-3Bx)/8=[7x-3*(-3)x]/8=[7x-(-9)x]/8=(7x+9x)/8=16x/8=2xAx=2xA=2

2006-10-28 18:20:34 補充：
NO.22(x^2-1)+Ax=(x+B)(Cx-1)L.H.S.=2(x^2-1)+Ax =2x^2-2+AxR.H.S.=(x+B)(Cx-1) =(Cx-1)x+(Cx-1)B =Cx^2-x+CBx-Bso, 2x^2-2+Ax=Cx^2-x+CBx-BBy comparing like terms,C=2B=-2Ax=CBx-x =2*(-2)*x-x =-4x-x =-5xAx=-5xA=-5

(8A+3B)x +2B≡7x-6
L.H.S. = (8A+3B)x +2B
R.H.S. = 7x-6
∴(8A+3B)x +2B≡7x-6
By the comparing of the like terms, we have,
2B=-6
B=-3
(8A+3B)=7
8A+3(-3)=7
8A-9=7
8A=16
A=16/8
A=2