Question

已知(x-y)^2=8,xy=2,求(x+y)^2,x^4+y^4的值

Answer 1

(x-y)^2=x^2+y^2-2xy=8

而xy=2 所以x^2+y^2=12

(x+y)^2=x^2+y^2+2xy=12+4=16

x^4+y^4=(x^2+y^2)^2-2x^2y^2

=12^2-2*2*2=144-8=136