已知x2+y2-2x+4y+5=0,求(x^4-y^4)/(2x^2+xy-y^2)·[(2x-y)/(xy-y^2)]÷[(x^2+y^2)/y]^2的值
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![已知x2+y2-2x+4y+5=0,求(x^4-y^4)/(2x^2+xy-y^2)·[(2x-y)/(xy-y^2)]÷[(x^2+y^2)/y]^2的值](/uploads/image/z/5432241-57-1.jpg?t=%E5%B7%B2%E7%9F%A5x2%2By2-2x%2B4y%2B5%3D0%2C%E6%B1%82%28x%5E4-y%5E4%29%2F%282x%5E2%2Bxy-y%5E2%29%C2%B7%5B%282x-y%29%2F%28xy-y%5E2%29%5D%C3%B7%5B%28x%5E2%2By%5E2%29%2Fy%5D%5E2%E7%9A%84%E5%80%BC)
已知x2+y2-2x+4y+5=0,求(x^4-y^4)/(2x^2+xy-y^2)·[(2x-y)/(xy-y^2)]÷[(x^2+y^2)/y]^2的值
已知x2+y2-2x+4y+5=0,求(x^4-y^4)/(2x^2+xy-y^2)·[(2x-y)/(xy-y^2)]÷[(x^2+y^2)/y]^2的值
已知x2+y2-2x+4y+5=0,求(x^4-y^4)/(2x^2+xy-y^2)·[(2x-y)/(xy-y^2)]÷[(x^2+y^2)/y]^2的值
x^2+y^2-2x+4y+5=0
(x-1)^2 + (y +2)^2 = 0
x= 1 ,y= -2
化解式子,再代入
x2+y2-2x+4y+5=0,则(x-1)²+(y+2)²=0,x=1,y=-2代入后式得(-15/-4)×(4/-6)÷(5/-2)²=3.75×(-2/3)÷6.25=-2.5÷6.25=-0.4