Đáp án:
a. \(\left[ \begin{array}{l}
x = 8\\
x = 2
\end{array} \right.\)
Giải thích các bước giải:
\(\begin{array}{l}
a.\left| {x - 5} \right| = 3\\
\to \left[ \begin{array}{l}
x - 5 = 3\\
x - 5 = - 3
\end{array} \right.\\
\to \left[ \begin{array}{l}
x = 8\\
x = 2
\end{array} \right.\\
b.\left| { - 5x} \right| = 3x - 16\\
\to \left[ \begin{array}{l}
- 5x = 3x - 16\\
5x = 3x - 16
\end{array} \right.\\
\to \left[ \begin{array}{l}
8x = 16\\
2x = - 16
\end{array} \right.\\
\to \left[ \begin{array}{l}
x = 2\\
x = - 8
\end{array} \right.\\
c.\left| {x - 4} \right| = - 3x + 5\\
\to \left[ \begin{array}{l}
x - 4 = - 3x + 5\\
x - 4 = 3x - 5
\end{array} \right.\\
\to \left[ \begin{array}{l}
4x = 9\\
2x = 1
\end{array} \right.\\
\to \left[ \begin{array}{l}
x = \dfrac{9}{4}\\
x = \dfrac{1}{2}
\end{array} \right.\\
d.\left| {3x - 1} \right| = x + 2\\
\to \left[ \begin{array}{l}
3x - 1 = x + 2\\
3x - 1 = - x - 2
\end{array} \right.\\
\to \left[ \begin{array}{l}
2x = 3\\
4x = - 1
\end{array} \right.\\
\to \left[ \begin{array}{l}
x = \dfrac{3}{2}\\
x = - \dfrac{1}{4}
\end{array} \right.\\
e.\left| {8 - x} \right| = {x^2} + x\\
\to \left[ \begin{array}{l}
8 - x = {x^2} + x\\
8 - x = - {x^2} - x
\end{array} \right.\\
\to \left[ \begin{array}{l}
{x^2} + 2x - 8 = 0\\
{x^2} + 8 = 0\left( l \right)
\end{array} \right.\\
\to \left( {x - 2} \right)\left( {x + 4} \right) = 0\\
\to \left[ \begin{array}{l}
x = 2\\
x = - 4
\end{array} \right.
\end{array}\)
