对任意实数m,直线(m+2)x-(m+1)y-2(3+2m)=0与P(-2,2)的距离d恒小于4√2

距离d=|-2(2+m)-2(1+m)-2(3+2m)|/√[(2+m)^2+(1+m)^2]
=|-4-2m-2-2m-6-4m|/√(4+4m+m^2+1+2m+m^2)
=|8m+12|/√(2m^2+6k+5)

令k=d^2=(64m^2+192m+144)/(2m^2+6m+5)
64m^2+192m+144=2km^2+6mk+5k
(64-2k)m^2+(192-6k)m+144-5k=0
这个关于m的方程有解
所以判别式=(192-6k)^2-4(64-2k)(144-5k)>=0
36864-2304k+36k^2-36864+2432k-40k^2>=0
k^2-32k<=0
0<=k<=32
0<=d^2<=32
0<=d<=4根号2

所以无论m取什么值，P到直线距离不大于 4根号2