题目内容
(1)计算:(-0.12)0+(
)-2•(3
)
-6•3-
+
(2)已知a+b=lg32+lg35+3lg2lg5,求a3+b3+3ab的值.
| 3 |
| 2 |
| 3 |
| 8 |
| 1 |
| 3 |
| 2 |
| 3 |
| 4 | 3
| |||
(2)已知a+b=lg32+lg35+3lg2lg5,求a3+b3+3ab的值.
(1)(-0.12)0+(
)-2•(3
)
-6•3-
+
=1+
×
-6•
+
=
-2•
+
=
-
(2)∵a+b=lg32+lg35+3lg2lg5
=(lg2+lg5)(lg22+lg25-lg2lg5)+3lg2lg5
=lg22+lg25-lg2lg5+3lg2lg5
=lg22+lg25+2lg2lg5
=(lg2+lg5)2=1
∴a3+b3+3ab=(a+b)(a2+b2-ab)+3ab
=a2+b2-ab+3ab=a2+b2+2ab=(a+b)2=1
| 3 |
| 2 |
| 3 |
| 8 |
| 1 |
| 3 |
| 2 |
| 3 |
| 4 | 3
| |||
=1+
| 4 |
| 9 |
| 3 |
| 2 |
| 1 | |||
|
| 3 | 3 |
=
| 5 |
| 3 |
| 3 | 3 |
| 3 | 3 |
=
| 5 |
| 3 |
| 3 | 3 |
(2)∵a+b=lg32+lg35+3lg2lg5
=(lg2+lg5)(lg22+lg25-lg2lg5)+3lg2lg5
=lg22+lg25-lg2lg5+3lg2lg5
=lg22+lg25+2lg2lg5
=(lg2+lg5)2=1
∴a3+b3+3ab=(a+b)(a2+b2-ab)+3ab
=a2+b2-ab+3ab=a2+b2+2ab=(a+b)2=1
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