Ta có: $|x+1| +|2x+1|+|3x+1|\geq \left | x+1+2x+1+3x+1 \right |=3(2x+1)$
mà $|x+1| +|2x+1|+|3x+1|=8x$
$\Rightarrow 3(2x+1)=8x$
$\Leftrightarrow 6x+3=8x$
$\Leftrightarrow 6x-8x=-3$
$\Leftrightarrow -2x=-3$
$\Leftrightarrow x=\frac{3}{2}$
Vậy $x=\frac{3}{2}$
Ta có: $|x-2| +|x-7|\geq \left | x-2+x-7 \right |=\left | 2x-9 \right |$
mà $|x-2| +|x-7|=5$
$\Rightarrow \left | 2x-9 \right |=5$
$\Leftrightarrow \left[ \begin{array}{l}2x-9=5\\2x-9=-5\end{array} \right.$
$\Leftrightarrow \left[ \begin{array}{l}x=7\\x=2\end{array} \right.$
Vậy $x\in \left \{ 7;2 \right \}$
