Question

已知x^2+4y^2-4x+4y+5=0,求(x^4-y^4/2x^2+xy-y^2)*(2x-y/xy-y^2)/(x

Answer 1

x^2+4y^2-4x+4y+5

=x^2-4x+4+4y^2+4y+1

=(x-2)^2+(2y+1)^2

=0

所以 x-2=0,2y+1=0

得 x=2,y= -1/2

代入后得y / (x^2+y^2)

结果为-2/17