F.2 maths (恆等式)~~~~~~~~~~20點

Cx^2+A(x+1)-B≡2(x+3)+3(x-2)
Left Hand Side:
Cx^2+Ax+A-B
=Cx^2+Ax+(A-B)
Right Hand Side:
2(x+3)+3(x-2)
=2x+6+3x-6
=2x+3x
=5x
=0x^2+5x+0
LHS and RHS:
Cx^2+Ax+(A-B)=0x^2+5x+0
Comparing x^2 term:
Cx^2=0x^2
C=0
Comparing x term:
Ax=5x
A=5
Comparing constant term:
(A-B)=0
A=B
Since A=5 and A=B
B=5
ANSWER:
A=5,B=5,C=0

Cx ^2 + A ( x + 1) - B ≡ 2 ( x + 3 ) + 3 ( x - 2 )
Cx ^2 + Ax + (A - B) ≡ 2x + 6 + 3x -6
Cx ^2 + Ax + (A - B) ≡ 5x
為右方加上 0x^2 和 0 ( 不會影響答案 )
Cx^2 + Ax + (A - B) ≡ 0x^2 + 5x + 0

比較同類項,
Cx^2 = 0x^2
C = 0

Ax = 5x
A = 5

(A-B) = 0
5-B=0
B= 5

所以 A = 5 , B=5, C=0

Cx ^2 + A ( x + 1) - B ≡ 2 ( x + 3 ) + 3 ( x - 2 )

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

RHS=2 ( x + 3 ) + 3 ( x - 2 )
=2x+6+3x-6
=5x

Comparing the like terms in LHS and RHS
A=5
C=0 (because RHS =0x^2+5x+0)

2010-09-08 20:28:41 補充：
A-B=0
5-B=0
B=5

Cx ^2 + A ( x + 1) - B ≡ 2 ( x + 3 ) + 3 ( x - 2 )
_________________________________________
LHS:
Cx ^2 + A ( x + 1) - B
Cx^2+Ax+A-B
Cx^2+Ax+(A-B)
_______________

RHS:
2 ( x + 3 ) + 3 ( x - 2 )
2x+6+3x-6
5x
__________________
Compare the like terms:
A=5

C=0

A-B=0
B=5

2010-09-08 21:01:54 補充：
No x^2 is on the question,so C is 0