题目内容
计算:(1)
x+2y |
x2-y2 |
y |
y2-x2 |
2x |
x2-y2 |
(2)
4 |
1 | ||
|
(3)
x2+2x+1 |
x+2 |
x2-1 |
x-1 |
1 |
x+2 |
(4)(
x |
x-1 |
2x |
x2-1 |
x |
x-1 |
分析:(1)先通分,再化简;
(2)根据二次根式、0指数幂、乘方、负指数幂的运算法则计算;
(3)分子分母能分解因式的分解,把除法转化为乘法,再约分,最后算减法;
(4)把除法转化为乘法,再按乘法分配律做.
(2)根据二次根式、0指数幂、乘方、负指数幂的运算法则计算;
(3)分子分母能分解因式的分解,把除法转化为乘法,再约分,最后算减法;
(4)把除法转化为乘法,再按乘法分配律做.
解答:解:(1)原式=
-
-
=
=
=-
;
(2)原式=2-1-8÷
=2-1-32
=-31;
(3)原式=
-
=
;
(4)原式=1-
•
=1-
=
=
.
故答案为-
、-31、
、
.
x+2y |
x2-y2 |
y |
x2-y2 |
2x |
x2-y2 |
=
x+2y-y-2x |
x2-y2 |
=
-x+y |
(x+y)(x-y) |
=-
1 |
x+y |
(2)原式=2-1-8÷
1 |
4 |
=2-1-32
=-31;
(3)原式=
x+1 |
x+2 |
1 |
x+2 |
=
x |
x+2 |
(4)原式=1-
2x |
(x+1)(x-1) |
x-1 |
x |
=1-
2 |
x+1 |
=
x+1-2 |
x+1 |
=
x-1 |
x+1 |
故答案为-
1 |
x+y |
x |
x+2 |
x-1 |
x+1 |
点评:(1)(3)(4)主要考查了分式的运算;(2)主要掌握二次根式、0指数幂、乘方、负指数幂的运算法则.
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