你好!
解:∵2x(y-1)+4(1/2x-1)=0
∴xy=2
∵(x+2y)(x-2y)=-5(y^2-6/5)
∴x²+y²=6
所以:
(1)(x-y)²=x²+y²-2xy=6-2*2=2
(2)x^4+y^4-x^2y^2
=(x²+y²)²-3(xy)²
=6²-3*2²=24
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∵2x(y-1)+4(1/2x-1)=0
∴xy=2
∵(x+2y)(x-2y)=-5(y^2-6/5)
∴x²+y²=6:
(1)(x-y)²=x²+y²-2xy=6-2*2=2
(2)x^4+y^4-x^2y^2
=(x²+y²)²-3(xy)²
=6²-3*2²=24