1、已知x^2+y^2-2x+4y+5=0,求4/(y+1/x)的值.

1个回答

  • 1、已知x^2+y^2-2x+4y+5=0,求4/(y+1/x)的值.

    (x-1)^2+(y+2)^2=0

    所以x-1=0,y+2=0

    x=1 y=-2

    y+1/x=-1

    4/(y+1/x)=-4

    2、已知x=1-a/1+a,y=3-2a/2-3a,用含x的代数式表示y.

    x=(1-a)/(1+a)

    x+xa=1-a

    a=(1-x)/(1+x)

    y=(2-3a)/(3-2a)

    3y-2ay=2-3a

    a=(2-3y)/(3-2y)

    (1-x)/(1+x)=(2-3y)/(3-2y)

    (1-x)(3-2y)=(2-3y)(1+x)

    3-3x-2y+2xy=2-3y+2x-3xy

    1+5xy=-y+5x

    5xy+y=5x-1

    y=(5x-1)/(5x+1)