Question

\(Tìm GTNN của A= \dfrac{(x+3)^2}{x+1} Biết x>-1\)

Answer 1

\(A=\dfrac{\left(x+3\right)^2}{x+1}\). Vì \(\left(x+3\right)^2\ge0\) ; \(x+1>0\left(x>-1\right)\)

\(\Rightarrow A=\dfrac{\left(x+3\right)^2}{x+1}\ge0\)

\(A_{min}=0\Leftrightarrow x=-3\left(tmđk\right)\)

Answer 2

\(A=\dfrac{x^2+6x+9}{x+1}=\dfrac{x^2-2x+1+8\left(x+1\right)}{x+1}=\dfrac{\left(x-1\right)^2}{x+1}+8\ge8\)

\(A_{min}=8\) khi \(x=1\)