题目内容
化简
(1)
lg25+lg2-lg
-log29×log32;
(2)
•b-2(
•b)-1÷(4-1a
b-3)
×
.
(1)
| 1 |
| 2 |
| 0.1 |
(2)
| 5 |
| 6 |
| 3 | a |
| a |
| 2 |
| 3 |
| 1 |
| 2 |
| ab |
(1)
lg25+lg2-lg
-log29×log32
=lg5+lg2+
-2
=-
.
(2)
•b-2(
•b)-1÷(4-1a
b-3)
×
=
×4×a
-
-
+
×b-2-1+
+
=
.
| 1 |
| 2 |
| 0.1 |
=lg5+lg2+
| 1 |
| 2 |
=-
| 1 |
| 2 |
(2)
| 5 |
| 6 |
| 3 | a |
| a |
| 2 |
| 3 |
| 1 |
| 2 |
| ab |
=
| 5 |
| 6 |
| 1 |
| 3 |
| 1 |
| 2 |
| 1 |
| 3 |
| 1 |
| 2 |
| 3 |
| 2 |
| 1 |
| 2 |
=
| 10 |
| 3b |
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