Ta có: \(\left|x+2\right|=\left|\frac{3}{2}x-1\right|\)
\(\Leftrightarrow\left[{}\begin{matrix}x+2=\frac{3}{2}x-1\\x+2=1-\frac{3}{2}x\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x+2-\frac{3}{2}x+1=0\\x+2-1+\frac{3}{2}x=0\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}\frac{-1}{2}x+3=0\\\frac{5}{2}x+1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}\frac{-1}{2}x=-3\\\frac{5}{2}x=-1\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-3:\frac{-1}{2}=6\\x=-1:\frac{5}{2}=-\frac{2}{5}\end{matrix}\right.\)
Vậy: \(x\in\left\{6;\frac{-2}{5}\right\}\)
Các bạn giúp mk vs, mk cần gấp, thanks👍