Đặt \(A=\left|2x-3\right|+2\left|x-1\right|\)
\(\Rightarrow A=\left|2x-3\right|+\left|2x-2\right|=\left|2x-3\right|+\left|2-2x\right|\)
\(\Rightarrow A\ge\left|2x-3+2-2x\right|=\left|-1\right|=1\)
Dấu " = " xảy ra \(\Leftrightarrow\left(2x-3\right)\left(2-2x\right)\ge0\)\(\Leftrightarrow\left(2x-3\right)\left(1-x\right)\ge0\)
TH1: \(\hept{\begin{cases}2x-3\le0\\1-x\le0\end{cases}}\Leftrightarrow\hept{\begin{cases}x\le\frac{3}{2}\\1\le x\end{cases}}\Leftrightarrow\hept{\begin{cases}x\le\frac{3}{2}\\x\ge1\end{cases}}\Leftrightarrow1\le x\le\frac{3}{2}\)
TH2: \(\hept{\begin{cases}2x-3\ge0\\1-x\ge0\end{cases}}\Leftrightarrow\hept{\begin{cases}x\ge\frac{3}{2}\\1\ge x\end{cases}}\Leftrightarrow\hept{\begin{cases}x\ge\frac{3}{2}\\x\le1\end{cases}}\)( vô lý )
Vậy \(minA=1\Leftrightarrow1\le x\le\frac{3}{2}\)
