题目内容
计算:
(1)(2
)0+2-2×(2
)-
-(0.01)0.5;
(2)lg14-21g
+lg7-lg18.
(1)(2
3 |
5 |
1 |
4 |
1 |
2 |
(2)lg14-21g
7 |
3 |
(1)(2
)0+2-2×(2
)-
-(0.01)0.5
=1+
×
-0.1
=1+
-
=
.
(2)lg14-21g
+lg7-lg18
=lg(14÷
×7÷18)
=lg1
=0.
3 |
5 |
1 |
4 |
1 |
2 |
=1+
1 |
4 |
2 |
3 |
=1+
1 |
6 |
1 |
10 |
=
16 |
15 |
(2)lg14-21g
7 |
3 |
=lg(14÷
49 |
9 |
=lg1
=0.
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