x-y=3,x^2+y^2=13 求xy和 x^4+y^4=

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x-y=3,x^2+y^2=13 求xy和 x^4+y^4=
x-y=3,x^2+y^2=13 求xy和 x^4+y^4=

x-y=3,x^2+y^2=13 求xy和 x^4+y^4=
x-y=3,x^2+y^2=13
(x-y)^2=9
x^2-2xy+y^2=9
-2xy=9-(x^2+y^2)=9-13=-4
xy=2
x^4+y^4
=(x^2+y^2)^2-2(xy)^2
=13^2-2*2^2
=169-8
=161

x-y=3
则 平方得
x^2-2xy+y^2=9
x^2+y^2=13
xy=2
x^2+y^2=13
平方得
x^4+2x^2y^2+y^4=169
xy=2
x^2y^2=4
x^4+y^4
=169-8
=161

（x-y)^2=9,xy=(x^2+y^2-9)/2=2 (x^2+y^2)^2=169....x^4+y^4=(169-2*(xy)^2)=161

﹙x－y﹚²＝3²
x²－2xy＋y²＝9
13－2xy＝9
xy＝﹣2;
x^4＋y^4＝﹙x²＋y²﹚²－2x²y²＝13²－2×﹙－2﹚²＝161.

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