恆等式兩條(請解答)

(2x+5)^2 -Ax ≡ 4x^2+16x+(29-2B)
_________
LHS:
(2x+5)^2-Ax
4x^2+20x+25-Ax
4x^2+x(20-A)+25
__________
RHS:
4x^2+16x+(29-2B)
________
Compare the like terms:
29-2B=25
B=2

20-A=16
A=4

2010-10-20 17:57:02 補充：
Ax(x-1)+B(x+1)(x-1)+C(x+1)x ≡ x^2+4
________
LHS:
Ax^2-Ax+Bx^2-B+Cx^2+Cx
x^2(A+B+C)+(-A+C)-B
__________
Compare the like terms:
-A+C=0
C=A

B=-4

A+C+-4=1
2A=5
A=5/2

C=5/2

2010-10-20 17:57:19 補充：
.

2010-10-20 18:56:32 補充：
The answer of Q2 which was solved by 002 is wrong.
I have checked before.

2010-10-20 18:57:41 補充：
To LeiMichael ,
係唔係maths in action 2A,6A(identity),最後果題?

2010-10-24 11:24:24 補充：
.

1) (2x+5)^2 -Ax ≡ 4x^2+16x+(29-2B)
LHS = (2x+5)^2 -Ax
= (2x)^2+2(2x)(5)+5^2-Ax
= 4x^2+20x+25-Ax
= 4x^2+(20-A)x+25
RHS = 4x^2+16x+(29-2B)
= 4x^2+16x+29-2B
因為(2x+5)^2 -Ax ≡ 4x^2+16x+(29-2B) ,
所以4x^2+(20-A)x+25 ≡ 4x^2+16x+29-2B
By compare,
20-A = 16
A = 4
29-2B = 25
2B = 4
B = 2

所以A = 4, B =2

2) Ax(x-1)+B(x+1)(x-1)+C(x+1)x ≡ x^2+4
LHS = Ax(x-1)+B(x+1)(x-1)+C(x+1)x
= Ax^2-Ax+Bx^2-B+Cx^2+x
= Ax^2+Bx^2+Cx^2-Ax+x-B
= (A+B+C)x^2-(A+1)x-B
RHS = x^2+4
因為Ax(x-1)+B(x+1)(x-1)+C(x+1)x ≡ x^2+4
所以(A+B+C)x^2-(A+1)x-B ≡ x^2+4

By compare,
B = 4
A+1 = 0
A = -1
A+B+C = 0
-1+4+C = 0
C = -3
所以 A = -1. B = 4, C = -3