$a) 2 . |x−1| = 16$
$⇔ |x-1| = 8$
$⇒$ \(\left[ \begin{array}{l}x-1=8\\x-1=-8\end{array} \right.\)
$⇔$ \(\left[ \begin{array}{l}x=9\\x=-7\end{array} \right.\)
Vậy $x$ $∈$ `{9;-7}`
$b(-12)^2 . x = 56 - (-10.13.x)$
$⇔ 144 .x = 56 + 130x$
$⇔ 144x-130x = 56$
$⇔ 14x =56$
$⇔ x =4$
Vậy $x=4$