题目内容
先化简,再求值:
(1)(
-
)÷
,其中x=
+1
(2)(1+
)÷
,其中x=
+3
(3)
÷(x+2-
),其中x=
-4
(4)
÷(
-x-2),其中x=
-2.
(1)(
| x+1 |
| x2-x |
| x |
| x2-2x+1 |
| 1 |
| x |
| 2 |
(2)(1+
| x-3 |
| x+3 |
| 2x |
| x2-9 |
| 3 |
(3)
| 4-x |
| x-2 |
| 12 |
| x-2 |
| 3 |
(4)
| x-3 |
| 2x-4 |
| 5 |
| x-2 |
| 3 |
分析:(1)先把除法转化成乘法,再利用乘法的分配律进行计算,然后把x的值代入计算即可;
(2)先把除法转化成乘法,再进行约分,然后把x的值代入计算即可;
(3)先算括号里面的,再把除法转化成乘法,再进行约分,然后把x的值代入计算即可;
(4)先算括号里面的,再把除法转化成乘法,再进行约分,然后把x的值代入计算即可;
(2)先把除法转化成乘法,再进行约分,然后把x的值代入计算即可;
(3)先算括号里面的,再把除法转化成乘法,再进行约分,然后把x的值代入计算即可;
(4)先算括号里面的,再把除法转化成乘法,再进行约分,然后把x的值代入计算即可;
解答:解:(1)(
-
)÷
,
=
•x-
•x,
=
-
,
=
,
=-
,
把x=
+1代入原式=
=3+
;
(2)(1+
)÷
,
=
×
=x-3;
把x=
+3代入原式=
+3-3=
;
(3)
÷(x+2-
),
=
÷
=
×
=-
,
把x=
-4代入原式=-
=-
;
(4)
÷(
-x-2),
=
÷
=
×
=-
把x=
-2代入上式得:
原式=-
=
.
| x+1 |
| x2-x |
| x |
| x2-2x+1 |
| 1 |
| x |
=
| x+1 |
| x(x-1) |
| x |
| (x-1)2 |
=
| x+1 |
| x-1 |
| x2 |
| (x-1)2 |
=
| x2-1-x2 |
| (x-2)2 |
=-
| 1 |
| (x-2)2 |
把x=
| 2 |
| 1 | ||
(
|
| 2 |
(2)(1+
| x-3 |
| x+3 |
| 2x |
| x2-9 |
=
| 2x |
| x+3 |
| (x+3)(x-3) |
| 2x |
=x-3;
把x=
| 3 |
| 3 |
| 3 |
(3)
| 4-x |
| x-2 |
| 12 |
| x-2 |
=
| 4-x |
| x-2 |
| (x+4)(x-4) |
| x-2 |
=
| 4-x |
| x-2 |
| x-2 |
| (x+4)(x-4) |
=-
| 1 |
| x+4 |
把x=
| 3 |
| 1 | ||
|
| ||
| 3 |
(4)
| x-3 |
| 2x-4 |
| 5 |
| x-2 |
=
| x-3 |
| 2(x-2) |
| 9-x2 |
| x-2 |
=
| x-3 |
| 2(x-2) |
| x-2 |
| (3+x)(3-x) |
=-
| 1 |
| 2((3+x) |
把x=
| 3 |
原式=-
| 1 | ||
2(3+
|
1-
| ||
| 4 |
点评:此题考查了分式的化简求值,解答此题的关键是把分式化到最简,然后代值计算.
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