题目内容
计算下列各式的值:
(1)(
)-2+(1-
)0-(3
)
;
(2)log2.56.25+lg0.01+ln
.
(1)(
| 2 |
| 3 |
| 2 |
| 3 |
| 8 |
| 2 |
| 3 |
(2)log2.56.25+lg0.01+ln
| e |
分析:(1)根据有理数指数幂的运算性质,我们根据(am)n=amn,a-1=
及a0=1,易得到结论.
(2)根据对数的运算性质,我们由logaan=n,alogaN=N化简式子,即可得到结果.
| 1 |
| a |
(2)根据对数的运算性质,我们由logaan=n,alogaN=N化简式子,即可得到结果.
解答:解:(1)(
)-2+(1-
)0-(3
)
=
+1-(
)
=
+1-[(
)3]
=
-(
)2=1
(2)log2.56.25+lg0.01+ln
═log2.52.52+lg10-2+lne
=2-2+
=
| 2 |
| 3 |
| 2 |
| 3 |
| 8 |
| 2 |
| 3 |
| 1 | ||
(
|
| 27 |
| 8 |
| 2 |
| 3 |
| 9 |
| 4 |
| 3 |
| 2 |
| 2 |
| 3 |
| 13 |
| 4 |
| 3 |
| 2 |
(2)log2.56.25+lg0.01+ln
| e |
| 1 |
| 2 |
| 1 |
| 2 |
| 1 |
| 2 |
点评:本题考查的知识点有理数指数幂的化简求值,对数的运算性质,熟练掌握对数的运算性质和指数运算的性质是解答本题的关键.
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