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 \( – 3 \le x \le 2\) Bình luận
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 \( – 3 \le x \le 2\)