1: Ta có: |-3x|=x+5
\(\Leftrightarrow\left[{}\begin{matrix}-3x=x+5\left(x\le0\right)\\3x=x+5\left(x>0\right)\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}-3x-x=5\\3x-x=5\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}-4x=5\\2x=5\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{-5}{4}\left(nhận\right)\\x=\dfrac{5}{2}\left(nhận\right)\end{matrix}\right.\)
Vậy: \(S=\left\{-\dfrac{5}{4};\dfrac{5}{2}\right\}\)
2: Ta có: \(10-\left|x+1\right|=3x+5\)
\(\Leftrightarrow\left|x+1\right|=10-3x-5\)
\(\Leftrightarrow\left|x+1\right|=-3x+5\)
\(\Leftrightarrow\left[{}\begin{matrix}x+1=-3x+5\left(x\ge-1\right)\\-x-1=-3x+5\left(x< -1\right)\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x+3x=5-1\\-x+3x=5+1\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}4x=4\\2x=6\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=1\left(nhận\right)\\x=3\left(loại\right)\end{matrix}\right.\)
Vậy:S={1}
3: Ta có: \(\dfrac{2x+3}{-4}\ge\dfrac{4-x}{-3}\)
\(\Leftrightarrow\dfrac{-2x-3}{4}\ge\dfrac{x-4}{3}\)
Suy ra: \(3\left(-2x-3\right)\ge4\left(x-4\right)\)
\(\Leftrightarrow-6x-9-4x+16\ge0\)
\(\Leftrightarrow-10x\ge-7\)
hay \(x\le\dfrac{7}{10}\)
Vậy: S={x|\(x\le\dfrac{7}{10}\)}