Đáp án:
a. \(x = - \dfrac{4}{3}\)
Giải thích các bước giải:
\(\begin{array}{l}
a.\left| {x + 1} \right| = \left| {2x + 3} \right|\\
\to \left[ \begin{array}{l}
x + 1 = 2x + 3\left( {DK:x \ge - 1} \right)\\
x + 1 = - 2x - 3\left( {DK:x < - 1} \right)
\end{array} \right.\\
\to \left[ \begin{array}{l}
x = - 2\left( l \right)\\
3x = - 4
\end{array} \right.\\
\to x = - \dfrac{4}{3}\\
b.\left| {x - 5} \right| + 2x = \left| {3x + 1} \right|\\
\to {x^2} - 10x + 25 + 2.2x.\left( {x - 5} \right) + 4{x^2} = 9{x^2} + 6x + 1\\
\to 4{x^2} + 16x - 24 = 4{x^2} - 20x\\
\to 36x = 24\\
\to x = \dfrac{2}{3}\\
c.\left| {x + 3} \right| = \left| {5x - 1} \right|\\
\to \left[ \begin{array}{l}
x + 3 = 5x - 1\left( {x \ge - 3} \right)\\
x + 3 = - 5x + 1\left( {x < - 3} \right)
\end{array} \right.\\
\to \left[ \begin{array}{l}
4x = - 4\\
6x = - 2
\end{array} \right.\\
\to \left[ \begin{array}{l}
x = - 1\\
x = - \dfrac{1}{3}\left( l \right)
\end{array} \right.
\end{array}\)
