b) Ta có: \(\left|x+1\right|+\left|4x-3\right|=\left|5x-2\right|\)
\(\Leftrightarrow x+1+4x-3+2\cdot\left|x+1\right|\left|4x-3\right|=5x-2\)
\(\Leftrightarrow5x-2+2\cdot\left|x+1\right|\left|4x-3\right|=5x-2\)
\(\Leftrightarrow2\cdot\left|x+1\right|\left|4x-3\right|=0\)
\(\Leftrightarrow\left(x+1\right)\left(4x-3\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x+1=0\\4x-3=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-1\\4x=3\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-1\\x=\frac{3}{4}\end{matrix}\right.\)
Vậy: \(x\in\left\{-1;\frac{3}{4}\right\}\)
