Đáp án:
\[ - 3 \le x \le 2\]
Giải thích các bước giải:
Ta có:
\(\begin{array}{l}
\left| {x - 2} \right| + \left| {x + 3} \right| = 5\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\left( 1 \right)\\
TH1:\,\,\,x < - 3 \Rightarrow \left\{ \begin{array}{l}
x - 2 < 0\\
x + 3 < 0
\end{array} \right. \Rightarrow \left\{ \begin{array}{l}
\left| {x - 2} \right| = - \left( {x - 2} \right)\\
\left| {x + 3} \right| = - \left( {x + 3} \right)
\end{array} \right.\\
\left( 1 \right) \Leftrightarrow - \left( {x - 2} \right) - \left( {x + 3} \right) = 5\\
\Leftrightarrow - 2x - 1 = 5\\
\Leftrightarrow - 2x = 6\\
\Leftrightarrow x = - 3\,\,\,\,\left( {L,\,\,\,x < - 3} \right)\\
TH2:\,\,\, - 3 \le x \le 2 \Rightarrow \left\{ \begin{array}{l}
x - 2 \le 0\\
x + 3 \ge 0
\end{array} \right. \Rightarrow \left\{ \begin{array}{l}
\left| {x - 2} \right| = - \left( {x - 2} \right)\\
\left| {x + 3} \right| = x + 3
\end{array} \right.\\
\left( 1 \right) \Leftrightarrow - \left( {x - 2} \right) + \left( {x + 3} \right) = 5\\
\Leftrightarrow 5 = 5,\,\,\,\forall x\\
\Rightarrow - 3 \le x \le 2\\
TH3:\,\,\,\,x > 2 \Rightarrow \left\{ \begin{array}{l}
x - 2 > 0\\
x + 3 > 0
\end{array} \right. \Rightarrow \left\{ \begin{array}{l}
\left| {x - 2} \right| = x - 2\\
\left| {x + 3} \right| = x + 3
\end{array} \right.\\
\left( 1 \right) \Leftrightarrow \left( {x - 2} \right) + \left( {x + 3} \right) = 5\\
\Leftrightarrow 2x + 1 = 5\\
\Leftrightarrow x = 2\,\,\,\,\left( {L,\,\,\,x > 2} \right)
\end{array}\)
Vậy \( - 3 \le x \le 2\)
