题目内容
(a+b)2=100,ab=20,求:(1)a2+b2,(2)a2-b2的值.
(1)∵(a+b)2=100,ab=20,
∴a2+b2+2ab=100,
即a2+b2=60;
(2)∵a2+b2=60,
∴(a2+b2)2=3600;
∴(a2-b2)2
=(a2+b2)2-4a2b2
=3600-4×400
=2000,
∴a2-b2=±
=±20
.
故答案为:60;±20
.
∴a2+b2+2ab=100,
即a2+b2=60;
(2)∵a2+b2=60,
∴(a2+b2)2=3600;
∴(a2-b2)2
=(a2+b2)2-4a2b2
=3600-4×400
=2000,
∴a2-b2=±
| 2000 |
| 5 |
故答案为:60;±20
| 5 |
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