Question

timf gtnn A=\(\dfrac{x+12}{\sqrt{x}+2}\)

Answer 1

ĐKXĐ: \(x\ge0\)

\(A=\dfrac{x+12}{\sqrt{x}+2}=\dfrac{x-4+16}{\sqrt{x}+2}=\dfrac{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)+16}{\sqrt{x}+2}=\sqrt{x}-2+\dfrac{16}{\sqrt{x}+2}=\sqrt{x}+2+\dfrac{16}{\sqrt{x}+2}-4\ge2\sqrt{\left(\sqrt{x}+2\right).\dfrac{16}{\sqrt{x}+2}}-4=2.4-4=4\)Dấu "=" xảy ra \(\Leftrightarrow\sqrt{x}+2=\dfrac{16}{\sqrt{x}+2}\Leftrightarrow x=4\)

Vậy \(MinA=4\)