x>-1,求y=(x+5)(x+2)/(x+1)的最小值?
y=(x+5)(x+2)/(x+1)=(x^2+7x+10)/(x+1)
y(x+1)=x^2+7x+10
x^2+(7-y)x+10-y=0………………(1)
∵x∈R,∴其判别式△=（7-y）^2-4(10-y)=y^2-14y+49-40+4y
=y^2-10y+9=(y-1)(y-9)≥0
得y≤1或y≥9.
将y=9代入（1）式,得x^2-2x+1=(x-1)^2=0,于是得x=1>-1.
将y=1代入（1）式,得x^2+6x+9=(x+3)^2=0,于是得x=-3-1,∴当x=1时获得y的最小值ymin=9