题目内容
(1)计算:
-
+(2-
)0+(-
)-1
(2)先化简,再求值:
÷(x-1-
),其中x=
+1.
|
(1-
|
| 5 |
| 1 |
| 2 |
(2)先化简,再求值:
| x2-2x |
| x2-1 |
| 2x-1 |
| x+1 |
| 2 |
分析:(1)根据二次根式的化简,任何非0数的0次幂等于1,有理数的负整数指数次幂等于正整数指数次幂的倒数进行计算即可得解;
(2)把被除数的分子分母分解因式,除式通分进行加减并把除法运算转化为乘法运算,然后约分,再把x的值代入进行计算即可得解.
(2)把被除数的分子分母分解因式,除式通分进行加减并把除法运算转化为乘法运算,然后约分,再把x的值代入进行计算即可得解.
解答:解:(1)
-
+(2-
)0+(-
)-1
=
-(
-1)+1-2
=
-
+1+1-2
=-
;
(2)
÷(x-1-
)
=
÷
=
×
=
,
当x=
+1时,原式=
=
=
.
|
(1-
|
| 5 |
| 1 |
| 2 |
=
| ||
| 6 |
| 3 |
=
| ||
| 6 |
| 3 |
=-
5
| ||
| 6 |
(2)
| x2-2x |
| x2-1 |
| 2x-1 |
| x+1 |
=
| x(x-2) |
| (x+1)(x-1) |
| x2-1-2x+1 |
| x+1 |
=
| x(x-2) |
| (x+1)(x-1) |
| x+1 |
| x(x-2) |
=
| 1 |
| x-1 |
当x=
| 2 |
| 1 |
| x-1 |
| 1 | ||
|
| ||
| 2 |
点评:本题主要考查了分式的化简求值,此类题目先化简后求值可以是运算更加简便,难点在于需要对通分、分解因式、约分等知识点熟练掌握.
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