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