恒等式~!帮我求D问号

2008-06-17 2:43 am
(1) 若 2x+A=Bx-1 , 求A及B.

左式= ?

右式= ?

A= ? B= ?

(2) 若 4(2m-1)+A=Bm+1 , 求A及B.

左式=4(2m-1)+A
=8m-( ? )+( ? )
=8m+(-4+A)

右式= ?
B= ?
-4+A= ?
A= ?

回答 (2)

2008-06-17 3:01 am
✔ 最佳答案
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
2008-06-17 3:55 am
(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


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