已知x²+y²-8x-10y+41=0,求x⼀y-y⼀x的值

2025-05-15 21:24:44
推荐回答(2个)
回答1:

x²+y²-8x-10y+41=0
(x²-8x+16)+(y²-10y+25)=0
(x-4)²+(y-5)²=0
x-4=0,y-5=0
x=4,y=5
x/y-y/x
=4/5-5/4
=-9/20

回答2:

x²+y²-8x-10y+41=0,
x^2-8x+16+y^2-10y+25=0
(x-4)^2+(y-5)^2=0
平方都大于等于0
相加为0则各项均为0
所以x-4=0
y-5=0
所以x=4,y=5
原式=4/5-5/4
=-9/20