\(\left(a\right)\Leftrightarrow\left\{{}\begin{matrix}5x-3>-2\\5x-3< 2\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}5x>1\\5x< 5\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x>\dfrac{1}{5}\\x< 1\end{matrix}\right.\) x thuộc Z => vô nghiệm
\(\left(b\right)\Leftrightarrow\left[{}\begin{matrix}3x+1>4\\3x+1< -4\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}3x>3\\3x< -5\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x>1\\x< -\dfrac{5}{3}\end{matrix}\right.\) \(x\in Z\backslash\left\{-1;0;1\right\}\)
\(\left(c\right)\Leftrightarrow\left|x-4\right|=3-2x\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x\ge4\\x-4=3-2x\Rightarrow x=\dfrac{7}{3}\left(l\right)\end{matrix}\right.\\\left\{{}\begin{matrix}x< 4\\x-4=2x-3\Rightarrow x=-1\left(n\right)\end{matrix}\right.\end{matrix}\right.\)\(\left[{}\begin{matrix}3x=7;x=\dfrac{7}{3}\\-4+3=x;x=-1\end{matrix}\right.\)