题目内容
计算:
(1)9
÷
×
;
(2)
+
+
-
+
;
(3)(
-
+
)•
;
(4)2a
-
-6ab
(b≥0).
(1)9
|
| 3 |
| 2 |
|
| 1 |
| 2 |
2
|
(2)
|
|
| 0.125 |
| 6 |
| 32 |
(3)(
| ||
| 6 |
| 4 | ||
|
| 5 | ||||
|
| 6 |
(4)2a
| 3ab2 |
| b |
| 6 |
| 27a3 |
|
考点:二次根式的混合运算
专题:计算题
分析:(1)先化简每一个二次根式,然后把除法转化为乘法再进行计算;
(2)先化简每一个二次根式,然后进行合并同类二次根式;
(3)根据乘法分配律进行计算;
(4)先化简每一个二次根式,然后进行合并同类二次根式
(2)先化简每一个二次根式,然后进行合并同类二次根式;
(3)根据乘法分配律进行计算;
(4)先化简每一个二次根式,然后进行合并同类二次根式
解答:解:(1)9
÷
×
=
÷
×
,
=
×
×
,
=
;
(2)
+
+
-
+
=
+
+
-
+4
=
-
;
(3)(
-
+
)•
=
•
-
•
+
•
=
-
+15
+10
=16
+
;
(4)2a
-
-6ab
=2ab
-
-2ab
=-
.
|
| 3 |
| 2 |
|
| 1 |
| 2 |
2
|
=
3
| ||
| 5 |
3
| ||
| 10 |
| ||
| 3 |
=
3
| ||
| 5 |
| 10 | ||
3
|
| ||
| 3 |
=
2
| ||
| 3 |
(2)
|
|
| 0.125 |
| 6 |
| 32 |
=
2
| ||
| 3 |
| ||
| 2 |
| ||
| 4 |
| 6 |
| 2 |
=
19
| ||
| 4 |
| ||
| 3 |
(3)(
| ||
| 6 |
| 4 | ||
|
| 5 | ||||
|
| 6 |
=
| ||
| 6 |
| 6 |
| 4 | ||
|
| 6 |
| 5 | ||||
|
| 6 |
=
| 2 |
4
| ||
| 3 |
| 2 |
| 3 |
=16
| 2 |
26
| ||
| 3 |
(4)2a
| 3ab2 |
| b |
| 6 |
| 27a3 |
|
=2ab
| 3a |
| ab |
| 2 |
| 3a |
| 3a |
=-
| ab |
| 2 |
| 3a |
点评:该题目考查了二次根式的混合运算,关键是灵活化简二次根式,理解二次根式的乘法、除法法则是基础.
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