Giải thích các bước giải:
Ta có:
a,
\(\begin{array}{l}
\left| {a + 3} \right| = 2a - 3\,\,\,\,\,\,\,\,\,\,\left( 1 \right)\\
\Rightarrow 2a - 3 \ge 0 \Rightarrow a \ge \frac{3}{2}\\
\left( 1 \right) \Leftrightarrow \left[ \begin{array}{l}
a + 3 = 2a - 3\\
a + 3 = 3 - 2a
\end{array} \right.\\
\Leftrightarrow \left[ \begin{array}{l}
a = 6\left( {t/m} \right)\\
a = 0\left( L \right)
\end{array} \right. \Rightarrow a = 6
\end{array}\)
Vậy \(a = 6\)
b,
Ta có:
\(\begin{array}{l}
\left| {a - 1} \right| + \left| {a - 3} \right| = 3a - 10\,\,\,\,\left( 1 \right)\\
TH1:\,\,\,a < 1 \Rightarrow \left\{ \begin{array}{l}
a - 1 < 0\\
a - 3 < 0
\end{array} \right.\\
\left( 1 \right) \Leftrightarrow \left( {1 - a} \right) + \left( {3 - a} \right) = 3a - 10\\
\Leftrightarrow 5a = 13\\
\Leftrightarrow a = \frac{{13}}{5}\,\,\,\left( {L,\,\,a < 1} \right)\\
TH2:\,\,\,1 \le a \le 3 \Rightarrow \left\{ \begin{array}{l}
a - 1 \ge 0\\
a - 3 \le 0
\end{array} \right.\\
\left( 1 \right) \Leftrightarrow \left( {a - 1} \right) + \left( {3 - a} \right) = 3a - 10\\
\Leftrightarrow 3a = 12\\
\Leftrightarrow a = 4\,\,\,\left( {L,\,\,1 \le a \le 3} \right)\\
TH3:\,\,a > 3 \Rightarrow \left\{ \begin{array}{l}
a - 1 > 0\\
a - 3 > 0
\end{array} \right.\\
\left( 1 \right) \Leftrightarrow \left( {a - 1} \right) + \left( {a - 3} \right) = 3a - 10\\
\Leftrightarrow a = 6\left( {t/m} \right)
\end{array}\)
Vậy \(a = 6\)
