Đáp án:
\(x = - \dfrac{{11}}{3}\) hoặc \(x = 3\)
Giải thích các bước giải:
Trường hợp 1: \(x < - 2\), ta có:
\(\begin{array}{l}
x < - 2 \Rightarrow \left\{ \begin{array}{l}
x + 2 < 0\\
4 - x > 0
\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}
\left| {x + 2} \right| = - \left( {x + 2} \right)\\
\left| {4 - x} \right| = 4 - x
\end{array} \right.\\
2\left| {x + 2} \right| + \left| {4 - x} \right| = 11\\
\Leftrightarrow - 2.\left( {x + 2} \right) + \left( {4 - x} \right) = 11\\
\Leftrightarrow - 2x - 4 + 4 - x = 11\\
\Leftrightarrow - 3x = 11\\
\Leftrightarrow x = - \dfrac{{11}}{3}\,\,\,\,\left( {t/m} \right)
\end{array}\)
Trường hợp 2: \( - 2 \le x \le 4\), ta có:
\(\begin{array}{l}
- 2 \le x \le 4 \Rightarrow \left\{ \begin{array}{l}
x + 2 \ge 0\\
4 - x \ge 0
\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}
\left| {x + 2} \right| = x + 2\\
\left| {4 - x} \right| = 4 - x
\end{array} \right.\\
2\left| {x + 2} \right| + \left| {4 - x} \right| = 11\\
\Leftrightarrow 2.\left( {x + 2} \right) + \left( {4 - x} \right) = 11\\
\Leftrightarrow 2x + 4 + 4 - x = 11\\
\Leftrightarrow 8 + x = 11\\
\Leftrightarrow x = 3\,\,\,\,\left( {t/m} \right)
\end{array}\)
Trường hợp 3: \(x > 4\), ta có:
\(\begin{array}{l}
x > 4 \Rightarrow \left\{ \begin{array}{l}
x + 2 > 0\\
4 - x < 0
\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}
\left| {x + 2} \right| = x + 2\\
\left| {4 - x} \right| = - \left( {4 - x} \right) = x - 4
\end{array} \right.\\
2\left| {x + 2} \right| + \left| {4 - x} \right| = 11\\
\Leftrightarrow 2.\left( {x + 2} \right) + \left( {x - 4} \right) = 11\\
\Leftrightarrow 3x = 11\\
\Leftrightarrow x = \dfrac{{11}}{3}
\end{array}\)
\(x = \dfrac{{11}}{3}\) không thoả mãn \(x > 4\)
Vậy \(x = - \dfrac{{11}}{3}\) hoặc \(x = 3\)
