\(A=\frac{2x}{3}+\frac{x+3}{x-1}=\frac{2}{3}x+\frac{x-1}{x-1}+\frac{4}{x-1}=\frac{2}{3}\left(x-1\right)+\frac{4}{x-1}+\frac{5}{3}\)
Áp dụng BĐT Cô-si:
\(A=\frac{2}{3}\left(x-1\right)+\frac{4}{x-1}+\frac{5}{3}\ge2\sqrt{\frac{2}{3}\left(x-1\right).\frac{4}{x-1}}+\frac{5}{3}=\frac{4\sqrt{6}}{3}+\frac{5}{3}=\frac{4\sqrt{6}+5}{3}\)
\(minA=\frac{4\sqrt{6}+5}{3}\Leftrightarrow\frac{2}{3}\left(x-1\right)=\frac{4}{x-1}\Leftrightarrow x=1\pm\sqrt{6}\)
Mà \(x>1\Rightarrow minA=\frac{4\sqrt{6}+5}{3}\Leftrightarrow x=1+\sqrt{6}\)
