x^2+y^2-10x-12y+61=0
(x²－10x＋25)＋(y²－12y＋36)=0
(x－5)²＋(y－6)²=0
因为(x－5)²与(y－6)²分别≥0
要想(x－5)²＋(y－6)²=0,必然
(x－5)²=0
x－5=0
x=5
(y－6)²=0
y－6=0
y=6
x/y-y/x
=5／6－6／5
=﹣11／30
已知x^2+y^2-10x-12y+61=0,求x/y-y/x的值
x^2+y^2-10x-12y+61=0
(x²－10x＋25)＋(y²－12y＋36)=0
(x－5)²＋(y－6)²=0
因为(x－5)²与(y－6)²分别≥0
要想(x－5)²＋(y－6)²=0,必然
(x－5)²=0
x－5=0
x=5
(y－6)²=0
y－6=0
y=6
x/y-y/x
=5／6－6／5
=﹣11／30