题目内容
计算:(1)(-3| 3 |
| 8 |
| 2 |
| 3 |
| 1 |
| 2 |
| 2 |
| 3 |
| 2 |
(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)原式=(
)
+(10-2)-
-
+1
=
+10-(
+1)+1(4分)
=
+10-
(6分)
=
-
(7分)
(2)原式=log2.52.52+lg10-2+lne
+2×2 log23
=2-2+
+2×3 (5分)
=
(7分)
| -27 |
| 8 |
| 2 |
| 3 |
| 1 |
| 2 |
| 1 | ||
|
=
| 9 |
| 4 |
| 2 |
=
| 9 |
| 4 |
| 2 |
=
| 49 |
| 4 |
| 2 |
(2)原式=log2.52.52+lg10-2+lne
| 1 |
| 2 |
=2-2+
| 1 |
| 2 |
=
| 13 |
| 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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