Question

x+y=3,xy=-2,试求代数式（1)(x-y）²; (2) x的四次方+y的四次方的值。

Answer 1

(x-y）²=（x+y）²-4xy=9+8=17
x的四次方+y的四次方=（x²+y²）²-2x²y²=（（x+y）²-2xy）²-2x²y²=（9+4）²-2*4=161

Answer 2

（x-y)^2=x^2-2xy+y^2
x+y=3
(x+y)^2=x^+2xy+y^2=9
xy=-2
则，x^2+y^=13
所以（x-y)^2=13+4=17

x^4+y^4=(x^2)^2+(y^2)^2
=(x^2+y^2)^2-2x^2y^2
=[(x+y)^2-2xy]^2-2(xy)^2
=[(9+4)^2]-8
=169-8
=161

Answer 3

（1)(x-y）²
=x^2+y^2-2xy
=(x+y)^2-4xy
=3^2-4*(-2)
=9+8
=17
(2) x^4+y^4
=(x^2+y^2)^2-2x^2y^2
=[(x+y)^2-2xy]^2-2(xy)^2
=[3^2-2*(-2)]^2-2*(-2)^2
=(9+4)^2-8
=13^2-8
=169-8
=161

Answer 4

1、(x－y)²=x²－2xy＋y²=(x²＋2xy＋y²)－4xy=(x＋y)²－4xy=17；
2、x^4＋y^4=(x²＋y²)²－2(xy)²=17²－8=281