Đáp án:
\[x \in \left\{ { - 2;\,\,7} \right\}\]
Giải thích các bước giải:
Ta có:
\(\begin{array}{l}
\left| {x - 4} \right| + \left| {x - 1} \right| = 9\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\left( 1 \right)\\
TH1:\,\,\,\,x < 1 \Rightarrow \left\{ \begin{array}{l}
x - 1 < 0\\
x - 4 < 0
\end{array} \right. \Rightarrow \left\{ \begin{array}{l}
\left| {x - 1} \right| = - \left( {x - 1} \right)\\
\left| {x - 4} \right| = - \left( {x - 4} \right)
\end{array} \right.\\
\left( 1 \right) \Leftrightarrow - \left( {x - 4} \right) - \left( {x - 1} \right) = 9\\
\Leftrightarrow - 2x + 5 = 9\\
\Leftrightarrow x = - 2\,\,\,\,\,\left( {t/m\,\,\,\,x < 1} \right)\\
TH2:\,\,\,1 \le x \le 4 \Rightarrow \left\{ \begin{array}{l}
x - 1 \ge 0\\
x - 4 \le 0
\end{array} \right. \Rightarrow \left\{ \begin{array}{l}
\left| {x - 1} \right| = x - 1\\
\left| {x - 4} \right| = - \left( {x - 4} \right)
\end{array} \right.\\
\left( 1 \right) \Leftrightarrow - \left( {x - 4} \right) + \left( {x - 1} \right) = 9\\
\Leftrightarrow 3 = 9\,\,\,\,\left( L \right)\\
TH3:\,\,\,x > 4 \Rightarrow \left\{ \begin{array}{l}
x - 1 > 0\\
x - 4 > 0
\end{array} \right. \Rightarrow \left\{ \begin{array}{l}
\left| {x - 1} \right| = x - 1\\
\left| {x - 4} \right| = x - 4
\end{array} \right.\\
\left( 1 \right) \Leftrightarrow \left( {x - 4} \right) + \left( {x - 1} \right) = 9\\
\Leftrightarrow 2x - 5 = 9\\
\Leftrightarrow x = 7\,\,\,\,\,\left( {t/m} \right)
\end{array}\)
Vậy \(x \in \left\{ { - 2;\,\,7} \right\}\) là nghiệm của phương trình đã cho.
