\(A=\frac{9x}{2-x}+\frac{2}{x}\)
\(=\frac{9x}{2-x}+\frac{2-x}{x}+1\)
AD BĐT Cosi cho 2 số thực không âm ta có:
\(\frac{9x}{2-x}+\frac{2-x}{x}\ge2\sqrt{\frac{9x}{2-x}.\frac{2-x}{x}}=2\sqrt{9}=6\)
\(\Rightarrow A\ge6+1=7\)
Dấu "=" xảy ra \(\Leftrightarrow\frac{9x}{2-x}=\frac{2-x}{x}\Leftrightarrow x=\frac{1}{2}\)
Vậy \(A_{min}=7\Leftrightarrow x=\frac{1}{2}\)
A=\(\frac{9x}{2-x}+\frac{2-x}{x}+1\)
Áp đụng bđt cô-si
A \(\ge2\ \cdot3\ +1\ =7\ \)