已知x²+y²-2x+4y+5=0,求x²-9y²/x²+6xy+9y²的值

问题描述:

已知x²+y²-2x+4y+5=0,求x²-9y²/x²+6xy+9y²的值

x²+y²-2x+4y+5=0
x²-2x+1+y²+4y+4=0
(x-1)^2+(y+2)^2=0
x-1=0 y+2=0
x=1 y=-2
x²-9y²/x²+6xy+9y²
=(x+3y)(x-3y)/(x+3y)^2
=x-3y /(x+3y)
=(1+6)(1-6)
=-7/5

先算出x=1,y=-2

x²+y²-2x+4y+5=0(x-1)²+(y+2)²=0x-1=0 y+2=0x=1 y= -2(x²-9y²)/(x²+6xy+9y²)=(x-3y)(x+3y)/(x+3y)²=(x-3y)/(x+3y)=(1+6)/(1-6)= -7/5...