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:365 ℃ 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
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