题目内容

已知(x+y)2=7,(x﹣y)2=3.
求:(1)x2+y2的值;
(2)x4+y4的值;
(3)x6+y6的值.
解:(1)∵(x+y)2=7,(x﹣y)2=3,
x2+2xy+y2=7,x2﹣2xy+y2=3,
∴x2+y2=5,xy=1;
(2)x4+y4=(x2+y22﹣2x2y2
=25﹣2
=23;
(3)x6+y6=(x2+y2)(x4﹣x2y2+y4
=5×(23﹣1)
=110.
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