Question

用分离常数法求值域y=（x方+2x+2）/（x+1）

Answer 1

y=(x²+2x+2)/(x+1)

=[(x+1)²+1]/(x+1)

=x+1+[1/(x+1)]

≥2

当且仅当x+1=1/(x+1)时即x=0或x=-2时取到等号