题目内容
(1)先化简,再求值(1+
)÷
,其中a=-3
(2)已知x-3y=0,求
.(x-y)的值.
| 1 |
| a2-1 |
| a |
| a-1 |
(2)已知x-3y=0,求
| 2x+y |
| x2-2xy+y2 |
分析:(1)先算括号里面的加法,同时把除法变成乘法,再进行约分,最后把a=-3代入求出即可;
(2)求出x=3y,分解因式得出
×(x-y),约分得出
,代入求出即可.
(2)求出x=3y,分解因式得出
| 2x+y |
| (x-y)2 |
| 2x+y |
| x-y |
解答:(1)解:
原式=(
+
)×
,
=
×
,
=
,
将a=-3代入得:
原式=
=
=
;
(2)解:∵x-3y=0,
∴x=3y,
∴
.(x-y),
=
×(x-y),
=
,
=
,
=
,
=
.
原式=(
| a2-1 |
| a2-1 |
| 1 |
| a2-1 |
| a-1 |
| a |
=
| a2 |
| (a+1)(a-1) |
| a-1 |
| a |
=
| a |
| a+1 |
将a=-3代入得:
原式=
| a |
| a+1 |
| -3 |
| -3+1 |
| 3 |
| 2 |
(2)解:∵x-3y=0,
∴x=3y,
∴
| 2x+y |
| x2-2xy+y2 |
=
| 2x+y |
| (x-y)2 |
=
| 2x+y |
| x-y |
=
| 2×3y+y |
| 3y-y |
=
| 7y |
| 2y |
=
| 7 |
| 2 |
点评:本题考查了分式的加减、乘除法的运用,主要考查学生的运算能力,题目比较好,是一道具有一定的代表性的题目.
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