\(0< x< \dfrac{1}{2}\).
\(A=\dfrac{2-x}{1-2x}+\dfrac{1+2x}{3x}=\dfrac{3x+2\left(1-2x\right)}{1-2x}+\dfrac{\left(1-2x\right)+4x}{3x}=\dfrac{3x}{1-2x}+2+\dfrac{1-2x}{3x}+\dfrac{4}{3}=\left(\dfrac{3x}{1-2x}+\dfrac{1-2x}{3x}\right)+\dfrac{10}{3}\ge^{Caushy}2\sqrt{\dfrac{3x}{1-2x}.\dfrac{1-2x}{3x}}+\dfrac{10}{3}=2+\dfrac{10}{3}=\dfrac{16}{3}\)
- Dấu "=" xảy ra \(\Leftrightarrow1-2x=3x\Leftrightarrow x=\dfrac{1}{5}\)
- Vậy \(MinA=\dfrac{16}{3}\)
\(A=\dfrac{2-x}{1-2x}+\dfrac{1+2x}{3x}\)
\(A=\dfrac{3x\left(2-x\right)}{3x\left(1-2x\right)}+\dfrac{\left(1-2x\right)\left(1+2x\right)}{3x\left(1-2x\right)}\)
\(A=\dfrac{6x-3x^2+1-4x^2}{3x\left(1-2x\right)}=\dfrac{-7x^2+6x+1}{3x\left(1-2x\right)}\)
\(A=\dfrac{\left(x-1\right)\left(x+\dfrac{1}{7}\right)}{3x\left(1-2x\right)}\)
Có: \(\left\{{}\begin{matrix}\left\{{}\begin{matrix}x-1< 0\\x+\dfrac{1}{7}>0\end{matrix}\right.\\\left\{{}\begin{matrix}3x>0\\1-2x< 0\end{matrix}\right.\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}\left(x-1\right)\left(x+\dfrac{1}{7}\right)< 0\\3x\left(1-2x\right)< 0\end{matrix}\right.\)
\(\Rightarrow\dfrac{\left(x-1\right)\left(x+\dfrac{1}{7}\right)}{3x\left(1-2x\right)}>0\) hay \(\dfrac{\left(x-1\right)\left(x+\dfrac{1}{7}\right)}{3x\left(1-2x\right)}\ge1\)
Vậy GTNN của A là 1
