题目内容
计算:
(1)
-
(2)
•(
)2÷
(3)
÷
-
(4)
÷(x-
).
(1)
| x2+y2 |
| x-y |
| 2xy |
| x-y |
(2)
| 2m |
| 3n |
| 3n |
| p |
| mn |
| p2 |
(3)
| a |
| a-1 |
| a2-a |
| a2-1 |
| 1 |
| a-1 |
(4)
| x2-y2 |
| x |
| 2xy-y2 |
| x |
分析:(1)直接计算即可;
(2)先算乘方,再从左向右计算;
(3)先算除法,再算减法;
(4)先算括号里的,再算括号外的.
(2)先算乘方,再从左向右计算;
(3)先算除法,再算减法;
(4)先算括号里的,再算括号外的.
解答:解:(1)原式=
=x-y;
(2)原式=
×
=6p;
(3)原式=
×
-
=
-
=
;
(4)原式=
×
=
.
| (x-y)2 |
| x-y |
(2)原式=
| 6mn |
| p |
| p2 |
| mn |
(3)原式=
| a |
| a-1 |
| (a+1)(a-1) |
| a(a-1) |
| 1 |
| a-1 |
| a+1 |
| a-1 |
| 1 |
| a-1 |
| a |
| a-1 |
(4)原式=
| (x+y)(x-y) |
| x |
| x |
| (x-y)2 |
| x+y |
| x-y |
点评:本题考查了分式的混合运算,解题的关键是注意分子、分母的因式分解,以及乘除法的转化.
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