Question

已知(2x²+2y²)(x²-2+y²)-16=0.求x²+y

Answer 1

(2x²+2y²)(x²-2+y²)-16=0

2(x²+y²)(x²+y²-2)=16

(x²+y²)(x²+y²-2)=8

把x²+y²看作一个整体,令a=x²+y²>0,则：

a(a-2)=8

a²-2a-8=0

(a-4)(a+2)=0

a=4；

a=-2