問中二數學

Does it want us to find A, B, C?

Ax(x-1)+B(x+1)(x-1)+C(x+1)x
=(A+B+C)x²-(A-C)x-B
=x²+4

Hence
B=-4
A=C

A+B+C=1
Putting B=-4
A=C

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

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

RHS (右) = x^2 + 4

左右比較 :
- B = 4 => B = - 4
C - A = 0 => A = C
A + B + C = 1 => 2A + B =1 => 2A - 4 = 1 => 2A = 5 => A= 2.5 = C

A = 2.5 B = - 4 C= 2.5

Ax(x-1)+B(x+1)(x-1)+C(x+1)x ≡ x²+4
Ax²-Ax+B(x²-1)+Cx+C ≡ x²+4
Ax²-Ax+Bx²-B+Cx+C ≡ x²+4
Ax²+Bx²-Ax+Cx-B+C≡ x²+4
x²(A+B)+x(C-A)+(C-B)≡ x²+4

C-A=0
C=A

A+B=1
B=1-A

C-(1-A)=4
C-1+A=4
C+A=5
C+C=5
C=2.5
A=2.5

B=1-A
B=-1.5