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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