已知x>4求y= -x+(1/4-x)的最大值

y=(4-x)+1/(4-x)-4
=-[(x-4)+1/(x-4)]-4
x>4,x-4>0
(x-4)+1/(x-4)>=2√[(x-4)*1/(x-4)]=2
所以-[(x-4)+1/(x-4)]<=-2
-[(x-4)+1/(x-4)]-4<=-6
所以最大值=-6

设M点的坐标为（x,y)
|MA|=|MB|
[(-3-x)^2+(2-y)^2]^(1/2)=[(3-x)^2+(-2-y)^2]^(1/2)
[(-3-x)^2+(2-y)^2]=[(3-x)^2+(-2-y)^2]
9+6x+x^2+4-4y+y^2=9-6x+x^2+4+4y+y^2
12x-8y=0
3x-2y=0
即证。