Ta có: \(\left|-2-5x\right|=-4x+7\)
=>\(\left|5x+2\right|=-4x+7\)
=>\(\begin{cases}-4x+7\ge0\\ \left(5x+2\right)^2=\left(-4x+7\right)^2\end{cases}\Rightarrow\begin{cases}-4x\ge-7\\ \left(5x+2+4x-7\right)\left(5x+2-4x+7\right)=0\end{cases}\)
=>\(\begin{cases}x\le\frac74\\ \left(9x-5\right)\left(x+9\right)=0\end{cases}\Rightarrow\begin{cases}x\le\frac74\\ x\in\left\lbrace\frac59;-9\right\rbrace\end{cases}\)
=>x∈{5/9;-9}