Question

Tìm GTNN của \(A=\frac{x+16}{\sqrt{x}+3}\)(x khác 1;x lớn hơn hoặc bằng 0)

Answer 1

\(A=\frac{x-9+25}{\sqrt{x}+3}=\frac{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}{\sqrt{x}+3}+\frac{25}{\sqrt{x}+3}=\sqrt{x}-3+\frac{25}{\sqrt{x}+3}\)

\(A=\left(\sqrt{x}+3\right)+\frac{25}{\sqrt{x}+3}-6\ge2.\sqrt{\left(\sqrt{x}+3\right).\frac{25}{\sqrt{x}+3}}-6=4\)

Dấu "=" xảy ra <=> \(\sqrt{x}+3=\frac{25}{\sqrt{x}+3}\) <=> \(\sqrt{x}+3=5\) <=> x = 4

Vậy....