题目内容
计算:
(1)
•(
)2-(
-
);
(2)
-
.
(1)
| x+1 |
| x |
| 2x |
| x+1 |
| 1 |
| x-1 |
| 1 |
| x+1 |
(2)
| 5x+3y |
| x2-y2 |
| 2x |
| x2-y2 |
分析:(1)根据分式混合运算的法则先算乘法,再算加减即可;
(2)根据同分母的分数相加减的法则进行计算即可.
(2)根据同分母的分数相加减的法则进行计算即可.
解答:解:(1)原式=
•
-
=
-
=
;
(2)原式=
=
=
.
| x+1 |
| x |
| 4x2 |
| (x+1)2 |
| x+1-x+1 |
| (x+1)(x-1) |
=
| 4x |
| x+1 |
| 2 |
| (x+1)(x-1) |
=
| 4x2-4x-2 |
| (x+1)(x-1) |
(2)原式=
| 5x+3y-2x |
| x2-y2 |
=
| 3(x+y) |
| (x+y)(x-y) |
=
| 3 |
| x-y |
点评:本题考查的是分式的混合运算,熟知分式混合运算的法则是解答此题的关键.
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