Đáp án:
d. \(\left[ \begin{array}{l}
x = \dfrac{7}{3}\\
x = - 5
\end{array} \right.\)
Giải thích các bước giải:
\(\begin{array}{l}
a.\left| {x - 5} \right| = 13 - 2x\\
\to \left[ \begin{array}{l}
x - 5 = 13 - 2x\left( {x \ge 5} \right)\\
x - 5 = - 13 + 2x\left( {x < 5} \right)
\end{array} \right.\\
\to \left[ \begin{array}{l}
3x = 18\\
x = 8\left( l \right)
\end{array} \right.\\
\to x = 6\left( {TM} \right)\\
b.\left| {5x - 1} \right| = x - 12\\
\to \left[ \begin{array}{l}
5x - 1 = x - 12\left( {x \ge \dfrac{1}{5}} \right)\\
5x - 1 = - x + 12\left( {x < \dfrac{1}{5}} \right)
\end{array} \right.\\
\to \left[ \begin{array}{l}
4x = - 11\\
6x = 13
\end{array} \right.\\
\to \left[ \begin{array}{l}
x = - \dfrac{{11}}{4}\left( l \right)\\
x = \dfrac{{13}}{6}\left( l \right)
\end{array} \right.
\end{array}\)
⇒ Phương trình vô nghiệm
\(\begin{array}{l}
c.\left| { - 2x} \right| = 3x + 4\\
\to \left[ \begin{array}{l}
- 2x = 3x + 4\left( {x \ge 0} \right)\\
2x = 3x + 4\left( {x < 0} \right)
\end{array} \right.\\
\to \left[ \begin{array}{l}
5x = - 4\\
x = - 4
\end{array} \right.\\
\to \left[ \begin{array}{l}
x = - \dfrac{4}{5}\left( l \right)\\
x = - 4\left( {TM} \right)
\end{array} \right.\\
d.\left| {2x - 1} \right| = 6 - x\\
\to \left[ \begin{array}{l}
2x - 1 = 6 - x\left( {x \ge \dfrac{1}{2}} \right)\\
2x - 1 = - 6 + x\left( {x < \dfrac{1}{2}} \right)
\end{array} \right.\\
\to \left[ \begin{array}{l}
3x = 7\\
x = - 5
\end{array} \right.\\
\to \left[ \begin{array}{l}
x = \dfrac{7}{3}\\
x = - 5
\end{array} \right.
\end{array}\)
