题目内容
计算:(1)
| x2 |
| (x+y)2 |
| 2y2 |
| (y+x)2 |
| y2-2xy |
| (-x-y)2 |
(2)
| 7 |
| x2+x |
| 3 |
| x2-x |
| 6 |
| x2-1 |
分析:(1)分式分母都相同,可以直接分式加减,(2)先确定最简公分母,再通分计算,直到结果为最简.
解答:解:(1)
+
-
=
=
=1,
(2)
+
-
=
+
-
=
=
=
.
| x2 |
| (x+y)2 |
| 2y2 |
| (y+x)2 |
| y2-2xy |
| (-x-y)2 |
=
| x2+2y2-y2+2xy |
| (x+y)2 |
=
| (x+y)2 |
| (x+y)2 |
=1,
(2)
| 7 |
| x2+x |
| 3 |
| x2-x |
| 6 |
| x2-1 |
=
| 7 |
| x(x+1) |
| 3 |
| x(x-1) |
| 6 |
| (x+1)(x-1) |
=
| 7x-7+3x+3-6x |
| x(x+1)(x-1) |
=
| 4(x-1) |
| x(x+1)(x-1) |
=
| 4 |
| x(x+1) |
点评:这两个题考查的分式的加减运算,(1)实质属于同分母分式的加减运算,应注意(x+y)2=(-x-y)2;(2)属于异分母分式的加减,关键是确定最简公分母.
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