\(a)\) \(2\left| {x - 1} \right| = 10 \)
\(\left| {x - 1} \right| = 5\)
Suy ra \(x + 1 = 5\) hay \(x = 4\)
hoặc \(x + 1 = -5\) hay \(x = -6\)
Vậy \(x = 4\) hoặc \(x = -6\)
\(b)\) \((-12)^2.x=56+10.13x\)
\(144x = 56 + 130x\)
\(14x = 56\)
Suy ra \(x = 4\)
