Đặt \(A=\frac{9x-2\sqrt{x}+4}{6\sqrt{x}}\)
\(A=\frac{9x}{6\sqrt{x}}-\frac{2\sqrt{x}}{6\sqrt{x}}+\frac{4}{6\sqrt{x}}\)
\(A=\frac{3\sqrt{x}}{2}+\frac{2}{3\sqrt{x}}-\frac{1}{3}\)
Áp dụng bất đẳng thức Cô-si:
\(A\ge2\sqrt{\frac{3\sqrt{x}\cdot2}{2\cdot3\sqrt{x}}}-\frac{1}{3}=2-\frac{1}{3}=\frac{5}{3}\)
Dấu "=" xảy ra \(\Leftrightarrow\frac{3\sqrt{x}}{2}=\frac{2}{3\sqrt{x}}\Leftrightarrow9x=4\Leftrightarrow x=\frac{4}{9}\)( thỏa )
Vậy....
\(DK:x>0\)
\(A=\frac{9x-2\sqrt{x}+4}{6\sqrt{x}}=\frac{3\sqrt{x}}{2}-\frac{1}{3}+\frac{2}{3\sqrt{x}}\)
\(\Rightarrow A+\frac{1}{3}=\frac{3\sqrt{x}}{2}+\frac{2}{3\sqrt{x}}\)
Áp dụng BĐT Cô-si
\(\Rightarrow A+\frac{1}{3}\ge2\sqrt{\frac{3\sqrt{x}}{2}.\frac{2}{3\sqrt{x}}}=2\sqrt{1}=2\)
\(\Rightarrow A\ge2-\frac{1}{3}=\frac{5}{3}\)
Vậy \(A_{min}=\frac{5}{3}\Leftrightarrow\frac{3\sqrt{x}}{2}=\frac{2}{3\sqrt{x}}\Leftrightarrow x=\frac{4}{9}\)
