Question

已知x²+y²-2x+4y+5=0，求（x四次方-y四次方）/（2x²+xy-y²）*（2x-y/

xy-y²）/（x²+y²/y）²

Answer 1

已知x²+y²-2x+4y+5=0，
x²-2x+1+y²+4y+4=0;
(x-1)²+(y+2)²=0;
x-1=0;
x=1;
y+2=0;
y=-2;
（x四次方-y四次方）/（2x²+xy-y²）*（2x-y/xy-y²）/（x²+y²/y）²
=（x-y）(x+y)(x²+y²)/(2x-y)（x+y）×(2x-y)/y(x-y)÷(x²+y²)²/y²
=(x-y)(x²+y²)/(2x-y)×(2x-y)y(x²+y²)²/(x-y)
=y(x²+y²)³
=-2×(1+4)³
=-2×125
=-250;