Đáp án:
c) \(x = - \dfrac{1}{2}\)
Giải thích các bước giải:
\(\begin{array}{l}
a)\dfrac{{3x + 5}}{2} - 1 \le \dfrac{{x + 2}}{3} + x\\
\to \dfrac{{9x + 15 - 6 - 2x - 4 - 6x}}{6} \le 0\\
\to x + 5 \le 0\\
\to x \le - 5\\
b)\left| {x - 3} \right| = 2x + 1\\
\to \left[ \begin{array}{l}
x - 3 = 2x + 1\left( {DK:x \ge 3} \right)\\
x - 3 = - 2x - 1\left( {DK:x < 3} \right)
\end{array} \right.\\
\to \left[ \begin{array}{l}
x = - 4\left( l \right)\\
3x = 2
\end{array} \right.\\
\to x = \dfrac{2}{3}\left( {TM} \right)\\
c)\left| {x + 3} \right| = \left| {3x + 4} \right|\\
\to \left[ \begin{array}{l}
x + 3 = 3x + 4\left( {DK:x \ge - 3} \right)\\
x + 3 = - 3x - 4\left( {DK:x < - 3} \right)
\end{array} \right.\\
\to \left[ \begin{array}{l}
2x = - 1\\
4x = - 7
\end{array} \right.\\
\to \left[ \begin{array}{l}
x = - \dfrac{1}{2}\left( {TM} \right)\\
x = - \dfrac{7}{4}\left( l \right)
\end{array} \right.
\end{array}\)
