用分离常数法求值域y=(x方+2x+2)/(x+1)
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用分离常数法求值域y=(x方+2x+2)/(x+1)
用分离常数法求值域
y=(x方+2x+2)/(x+1)
用分离常数法求值域y=(x方+2x+2)/(x+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
y=[(x+1)^2+1]/(x+1)
y=(x+1)+1/(x+1)≥2
所以值域为[2,正无穷)