实数x,y满足x2+y2-2x-2y+1=0,求y-4⼀x-2的取值范围

2025-05-16 23:06:11
推荐回答(1个)
回答(1):

首先配方,将式子变成(x-1)^2+(y-1)^2=1,因为0<=(x-1)^2<=1,0<=(y-1)^2<=1,所以可以得出
1<=x<=2,1<=y<=2,然后
-1<=x-2<=0,-3<=y-4<=-2

所以y-4/x-2的取值范围
0到1/2