题目内容
求下列函数的值:
(1)(
)
×(
)-1+4×(
)-
;
(2)lg
-
lg2
+lg7
.
(1)(
| 9 |
| 25 |
| 1 |
| 2 |
| 1 |
| 10 |
| 8 |
| 27 |
| 2 |
| 3 |
(2)lg
4
| ||
| 7 |
| 4 |
| 3 |
| 3 |
| 2 |
| 5 |
分析:(1)利用有理数指数幂的性质,把(
)
×(
)-1+4×(
)-
等价转化为
×10+4×(
)-2,由此能够求出结果.
(2)利用对数的运算性质和运算法则,把lg
-
lg2
+lg7
等价转化为lg(
÷2
×
×7
),由此能够求出结果.
| 9 |
| 25 |
| 1 |
| 2 |
| 1 |
| 10 |
| 8 |
| 27 |
| 2 |
| 3 |
| 3 |
| 5 |
| 2 |
| 3 |
(2)利用对数的运算性质和运算法则,把lg
4
| ||
| 7 |
| 4 |
| 3 |
| 3 |
| 2 |
| 5 |
4
| ||
| 7 |
| 3 |
| 2 |
| 4 |
| 3 |
| 5 |
解答:解:(1)(
)
×(
)-1+4×(
)-
=
×10+4×(
)-2
=6+4×
=15.
(2)lg
-
lg2
+lg7
=lg(
÷2
×
×7
)
=lg
=
.
| 9 |
| 25 |
| 1 |
| 2 |
| 1 |
| 10 |
| 8 |
| 27 |
| 2 |
| 3 |
=
| 3 |
| 5 |
| 2 |
| 3 |
=6+4×
| 9 |
| 4 |
=15.
(2)lg
4
| ||
| 7 |
| 4 |
| 3 |
| 3 |
| 2 |
| 5 |
=lg(
4
| ||
| 7 |
| 3 |
| 2 |
| 4 |
| 3 |
| 5 |
=lg
| 10 |
=
| 1 |
| 2 |
点评:本题考查有理数指数幂的性质及其去处法则,考查对数的运算性质和运算法则,是基础题.解题时要认真审题,仔细解答.
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