Đáp án:
\[S = \left( { - \infty ; - 2} \right) \cup \left( {5; + \infty } \right)\]
Giải thích các bước giải:
\(\begin{array}{l}
\left| {x + 1} \right| + \left| {x - 4} \right| > 7\,\,\,\,\,\,\,\,\,\,\,\,\,\left( 1 \right)\\
TH1:\,\,x < - 1 \Rightarrow \left\{ \begin{array}{l}
x + 1 < 0\\
x - 4 < 0
\end{array} \right.\\
\left( 1 \right) \Leftrightarrow - \left( {x + 1} \right) - \left( {x - 4} \right) > 7\\
\Leftrightarrow 3 - 2x > 7\\
\Leftrightarrow 2x < - 4\\
\Leftrightarrow x < - 2\\
\Rightarrow {S_1} = \left( { - \infty ; - 2} \right)\\
TH2:\,\,\, - 1 \le x \le 4 \Rightarrow \left\{ \begin{array}{l}
x + 1 \ge 0\\
x - 4 \le 0
\end{array} \right.\\
\left( 1 \right) \Leftrightarrow \left( {x + 1} \right) - \left( {x - 4} \right) > 7\\
\Leftrightarrow 5 > 7\,\,\left( {vn} \right)\\
\Rightarrow {S_2} = \left\{ \emptyset \right\}\\
TH3:\,\,x > 4 \Rightarrow \left\{ \begin{array}{l}
x + 1 > 0\\
x - 4 > 0
\end{array} \right.\\
\left( 1 \right) \Leftrightarrow x + 1 + x - 4 > 7\\
\Leftrightarrow 2x > 10\\
\Leftrightarrow x > 5\\
\Rightarrow {S_3} = \left( {5; + \infty } \right)\\
\Rightarrow S = {S_1} \cup {S_1} \cup {S_3} = \left( { - \infty ; - 2} \right) \cup \left( {5; + \infty } \right)
\end{array}\)
