d) Ta có: \(3\cdot\left|x-1\right|-27=0\)
\(\Leftrightarrow3\cdot\left|x-1\right|=27\)
\(\Leftrightarrow\left|x-1\right|=9\)
\(\Leftrightarrow\left[{}\begin{matrix}x-1=9\\x-1=-9\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=10\\x=-8\end{matrix}\right.\)
Vậy: \(x\in\left\{-8;10\right\}\)
e) Ta có: \(5\cdot\left(3x+8\right)-7\cdot\left(2x+3\right)=16\)
\(\Leftrightarrow15x+40-14x-21=16\)
\(\Leftrightarrow x+19=16\)
hay x=-3
Vậy: x=-3