c) \(\left|x+3\right|-2x=\left|x-4\right|\)
\(\Rightarrow\left|x+3\right|-2x-\left|x-4\right|=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x+3-2x-\left(x-4\right)=0\left(đk:x+3\ge0,x-4\ge0\right)\\-\left(x+3\right)-2x-\left(x-4\right)=0\left(đk:x+3< 0,x-4\ge0\right)\\x+3-2x-\left(-\left(x-4\right)\right)=0\left(đk:x+3\ge0,x-4< 0\right)\\-\left(x+3\right)-2x-\left(-\left(x-4\right)\right)=0\left(đk:x+3< 0,x-4< 0\right)\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{7}{2}\left(đk:x\ge-3;x\ge4\right)\\x=\dfrac{1}{4}\left(đk:x< -3,x\ge4\right)\\x\in\varnothing\left(đk:x\ge-3,x< 4\right)\\x=-\dfrac{7}{2}\left(đk:x< -3;x< 4\right)\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x\in\varnothing\\x\in\varnothing\\x\in\varnothing\\x=-\dfrac{7}{2}\end{matrix}\right.\)
Vậy \(x=-\dfrac{7}{2}\)
