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^2+y^_数学解答

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^2+y^2/y)^2 x^3-2x-9

人气：486 ℃　时间：2019-10-17 01:29:44

解答

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