题目内容
计算:(1)
| 8 |
|
| 1 | ||
|
| ||
| 2 |
(2)
| 2 |
| b |
| ab5 |
| 3 |
| 2 |
| a3b |
|
(3)已知x=
| 2 |
| x+1 |
| x2-x |
| x |
| x2-2x+1 |
| 1 |
| x |
分析:(1)(2)根据二次根式的性质进行求解;
(3)先对式子进行化简,然后再把x=
+1代入(
-
)÷
求解.
(3)先对式子进行化简,然后再把x=
| 2 |
| x+1 |
| x2-x |
| x |
| x2-2x+1 |
| 1 |
| x |
解答:解:(1)
+3
-
+
=2
+
-
+
=
+
;
(2)
•(-
)÷3
=
×b2×
×(-
)×a
×
=-a2b
;
(3)(
-
)÷
=(
-
)×x
=-
∵x=
+1,∴x2=3+2
=-
=-
.
| 8 |
|
| 1 | ||
|
| ||
| 2 |
| 2 |
| 3 |
| ||
| 2 |
| ||
| 2 |
=
| 3 |
| 2 |
| 2 |
| 3 |
| 2 |
| 3 |
(2)
| 2 |
| b |
| ab5 |
| 3 |
| 2 |
| a3b |
|
| 2 |
| b |
| ab |
| 3 |
| 2 |
| ab |
| ||
3
|
=-a2b
| ab |
(3)(
| x+1 |
| x2-x |
| x |
| x2-2x+1 |
| 1 |
| x |
| x+1 |
| x(x-1) |
| x |
| (x-1)2 |
=-
| 1 |
| x2-1 |
∵x=
| 2 |
| 2 |
| 1 |
| x2-1 |
| 1 | ||
3+2
|
=-
| 1 |
| 2 |
点评:此题主要考查二次根式的性质和化简,计算时要仔细,是一道基础题.
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