x,y满足x²+y²=1,求(x+y+2)/(x-y+2)的最值.
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x,y满足x²+y²=1,求(x+y+2)/(x-y+2)的最值.
x,y满足x²+y²=1,求(x+y+2)/(x-y+2)的最值.
x,y满足x²+y²=1,求(x+y+2)/(x-y+2)的最值.
(x+y+2)/(x-y+2)=(x-y+2+2y)/(x-y+2)=1-[2y/(x-y+2)]分式上下同除2
得1-{2/[(x+2)/y-1]}
另x^2+y^2=1 (x,y)是以1为半径,原点为圆心的圆的轨迹
则上式中(x+2)/y=[x-(-2)]/(y-0)为圆上的点轨迹到(-2,0)点的直线斜率的倒数
设k=y/(x+2) 设经过点(-2,0)且切于圆的直线为y=kx+b,则有-2k+b=0
将y=kx+b代入x^2+y^2=1
得(k^2+1)x^2+2kbx+b^2-1=0
切于圆,有一个交点,则△=0
即k^2-b^2+1=0
将-2k+b=0代入上式
可得k=±√3/3
则1/k=±√3
将上值代入原式得最大值为1+2/(√3+1)最小值为1-2/(√3-1)