Question

Tìm GTNN của \(\frac{x+3\sqrt{x-1}+1}{x+4\sqrt{x-1}+2}\) với x lớn hơn hoặc = 1

Answer 1

\(A=\frac{x+3\sqrt{x-1}+1}{x+4\sqrt{x-1}+2}=\frac{x-1+3\sqrt{x-1}+2}{x-1+4\sqrt{x-1}+3}\)

\(=\frac{\left(\sqrt{x-1}+1\right)\left(\sqrt{x-1}+2\right)}{\left(\sqrt{x-1}+1\right)\left(\sqrt{x-1}+3\right)}\)

\(=\frac{\sqrt{x-1}+2}{\sqrt{x-1}+3}\)

\(=1-\frac{1}{\sqrt{x-1}+3}\)

Ta có : \(\sqrt{x-1}\ge0\)

\(\Leftrightarrow\sqrt{x-1}+3\ge3\)

\(\Leftrightarrow\frac{1}{\sqrt{x-1}+3}\le\frac{1}{3}\)

\(\Leftrightarrow1-\frac{1}{\sqrt{x-1}+3}\ge1-\frac{1}{3}=\frac{2}{3}\)

\(\Rightarrow A\ge\frac{2}{3}\)

\(\text{GTNN của A bằng }\frac{3}{2}\)

\(\text{Dấu }=\text{xảy ra }\Leftrightarrow x=1\)