Ta có :
\(\left|2x\left(x+3\right)\right|\ge0\)
\(\Rightarrow\)\(x\ge0\)
\(PT\)\(\Leftrightarrow\)\(2x\left(x+3\right)=x\)
\(\Leftrightarrow\)\(2x^2+6x=x\)
\(\Leftrightarrow\)\(2x^2+6x-x=0\)
\(\Leftrightarrow\)\(2x^2+5x=0\)
\(\Leftrightarrow\)\(x\left(2x+5\right)=0\)
\(\Leftrightarrow\)\(\orbr{\begin{cases}x=0\\2x+5=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=0\\x=\frac{-5}{2}\end{cases}}}\)
Vậy \(x=0\) hoặc \(x=\frac{-5}{2}\)
\(\left|2-3x\right|=\left|-\frac{1}{2}\right|\)
\(\Leftrightarrow\)\(\orbr{\begin{cases}2-3x=\frac{-1}{2}\\2-3x=\frac{1}{2}\end{cases}\Leftrightarrow\orbr{\begin{cases}3x=2+\frac{1}{2}\\3x=2-\frac{1}{2}\end{cases}}}\)
\(\Leftrightarrow\)\(\orbr{\begin{cases}3x=\frac{5}{2}\\3x=\frac{3}{2}\end{cases}\Leftrightarrow\orbr{\begin{cases}x=\frac{5}{6}\\x=\frac{1}{2}\end{cases}}}\)
Vậy \(x=\frac{5}{6}\) hoặc \(x=\frac{1}{2}\)
Chúc bạn học tốt ~
