(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= ?
恒等式~!帮我求D问号
2008-06-17 2:43 am
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
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
收錄日期: 2021-04-13 15:42:19
原文連結 [永久失效]:
https://hk.answers.yahoo.com/question/index?qid=20080616000051KK02192