题目内容
(1)求值:lg52+
lg8+lg5•lg20+(lg2)2
(2)求值:(0.0081)-
-[3×(
)0]-1×[81-0.25+(3
)-
]-
-10×0.027
.
| 2 |
| 3 |
(2)求值:(0.0081)-
| 1 |
| 4 |
| 7 |
| 8 |
| 3 |
| 8 |
| 1 |
| 3 |
| 1 |
| 2 |
| 1 |
| 3 |
分析:(1)把第二项真数上的8化为23,第三项中的真数上的20化为2×10,然后利用对数的运算性质化简求值;
(2)化小数为分数,化负指数为正指数,化带分数为假分数,然后进行有理指数幂的化简运算.
(2)化小数为分数,化负指数为正指数,化带分数为假分数,然后进行有理指数幂的化简运算.
解答:解:(1)lg52+
lg8+lg5•lg20+(lg2)2
=2lg5+
lg23+lg5(1+lg2)+(lg2)2
=2lg5+2lg2+lg5(1+lg2)+(lg2)2
=2(lg5+lg2)+lg5+lg5•lg2+(lg2)2
=2+lg5+lg2(lg5+lg2)=3.
(2)(0.0081)-
-[3×(
)0]-1×[81-0.25+(3
)-
]-
-10×0.027
=((0.3)4)-
-3-1×[(34)-
+(
)-1]-
-10×((0.3)3)
=
-
(
+
)-
-10•0.3
=
-
-3
=0.
| 2 |
| 3 |
=2lg5+
| 2 |
| 3 |
=2lg5+2lg2+lg5(1+lg2)+(lg2)2
=2(lg5+lg2)+lg5+lg5•lg2+(lg2)2
=2+lg5+lg2(lg5+lg2)=3.
(2)(0.0081)-
| 1 |
| 4 |
| 7 |
| 8 |
| 3 |
| 8 |
| 1 |
| 3 |
| 1 |
| 2 |
| 1 |
| 3 |
=((0.3)4)-
| 1 |
| 4 |
| 1 |
| 4 |
| 3 |
| 2 |
| 1 |
| 2 |
| 1 |
| 3 |
=
| 10 |
| 3 |
| 1 |
| 3 |
| 1 |
| 3 |
| 2 |
| 3 |
| 1 |
| 2 |
=
| 10 |
| 3 |
| 1 |
| 3 |
=0.
点评:本题考查了有理指数幂的化简求值,考查了对数式的运算性质,解答的关键是熟记有关公式,此题是基础题.
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