题目内容
(Ⅰ)计算π0+2-2×(
)-
-(0.01)0.5;
(Ⅱ)计算2log510+log50.25.
| 9 |
| 4 |
| 1 |
| 2 |
(Ⅱ)计算2log510+log50.25.
考点:对数的运算性质,根式与分数指数幂的互化及其化简运算
专题:函数的性质及应用
分析:(Ⅰ)利用分数指数幂的去处法则求解.
(Ⅱ)利用对数的去处法则求解.
(Ⅱ)利用对数的去处法则求解.
解答:
解:(Ⅰ)π0+2-2×(
)-
-(0.01)0.5;
=1+
×[(
)-1]
-0.01
=1+
×[
]
-(
)
=1+
×
-
=
.…(6分)
(Ⅱ)2log510+log50.25
=log5100+log50.25
=log525=2.
| 9 |
| 4 |
| 1 |
| 2 |
=1+
| 1 |
| 22 |
| 9 |
| 4 |
| 1 |
| 2 |
| 1 |
| 2 |
=1+
| 1 |
| 4 |
| 4 |
| 9 |
| 1 |
| 2 |
| 1 |
| 100 |
| 1 |
| 2 |
=1+
| 1 |
| 4 |
| 2 |
| 3 |
| 1 |
| 10 |
| 16 |
| 15 |
(Ⅱ)2log510+log50.25
=log5100+log50.25
=log525=2.
点评:本题考查指数式和对数式化简求值,是基础题,解题时要注意运算法则的合理运用.
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