恒等式~!帮我求D问号

1

既然是恆等式

左式
= 2(x) + (A) (x的coefficient = 2 constant = A)

右式
= B(x) + (-1) (x的coefficient = B constant = -1)

所以

A = -1 (constant)
B = 2 (x的coefficient)

2
同理

已給了左式
4(2m-1)+A
= 8m -4 + A
= 8m+(-4+A) (m的coefficient = 8 constant = -4+A)

右式

= B(m) + (1) (m的coefficient = B constant = 1)

所以

-4+A = 1
A = 5
B = 8

(1)
A=-1
B=2
左式=2x-1
右式=2x-1

(2)
左式=4(2m-1)+A
　　=8m-4+A
　　=8m+(-4+A)
右式=Bm-1
B=8
A=3