Question

已知x^2+4y^2-2x-2x+8y+5=0,求x^2-y^2/2x^2+xy-y^2*2y-y/xy-y/(x^2+y^2/y)^2的值

Answer 1

x²+4y²-2x+8y+5=0
(x-1)²+(2y+2)²=0
得 x=1，y=-1
则 x²-y²/2x²+xy-y²*2y-y/xy-y/(x²+y²/y)²
=1-1/2-1+2-1+1/4
=1/2+1/4
=3/4

追问

已知x^2+4y^2-2x-2x+8y+5=0,求x^4-y^4/2x^2+xy-y^2*2y-y/xy-y/(x^2+y^2/y)^2的值

追答

x^2+4y^2-2x+8y+5=0
(x-1)²+(2y+2)²=0
得 x=1，y=-1
则 x^4-y^4/2x^2+xy-y^2*2y-y/xy-y/(x^2+y^2/y)^2

=1-1/2-1+2-1+1/4
=3/4

Answer 2

原题是多打了一个2x吗？。。如果没有那个2x，那么原式可以化为：(x-1)^2+4(y+1)^2=0，然后就x=1 y=-1.