已知x2+y2-2x-6y+10=0,则4/(y+1/x)=
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x^2+y^2-2x-6y+10=0
x^2-2x+4+y^2-6y+6=0
(x-2)^2+(y-3)^2=0
x-2=0 x=2
y-3=0 y=3
4/[y+(1/x)]
=4/[3+(1/2)]
=4/(7/2)
=8/7
4/[(y+1)/x]
=4/(3+1)/2
=4/2
=2
==========
后面你写的不明确
x^2-2x+4+y^2-6y+6=0
(x-2)^2+(y-3)^2=0
x-2=0 x=2
y-3=0 y=3
4/[y+(1/x)]
=4/[3+(1/2)]
=4/(7/2)
=8/7
4/[(y+1)/x]
=4/(3+1)/2
=4/2
=2
==========
后面你写的不明确
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