Đáp án:
\[S = \left( { - \frac{{16}}{3}; - \frac{2}{3}} \right)\]
Giải thích các bước giải:
Ta có:
\(\begin{array}{l}
\frac{{\left| {4 - x} \right| + x + 3}}{{\left| {x + 3} \right|}} > 3\\
\Leftrightarrow \left| {4 - x} \right| + x + 3 - 3\left| {x + 3} \right| > 0\,\,\,\,\,\,\,\left( 1 \right)
\end{array}\)
Xét các TH sau:
\(\begin{array}{l}
TH1:\,\,\,x < - 3 \Rightarrow \left\{ \begin{array}{l}
4 - x > 0\\
x + 3 < 0
\end{array} \right. \Rightarrow \left\{ \begin{array}{l}
\left| {4 - x} \right| = 4 - x\\
\left| {x + 3} \right| = - \left( {x + 3} \right)
\end{array} \right.\\
\left( 1 \right) \Leftrightarrow \left( {4 - x} \right) + x + 3 + 3\left( {x + 3} \right) > 0\\
\Leftrightarrow 3x + 16 > 0\\
\Leftrightarrow x > - \frac{{16}}{3}\\
\Rightarrow {S_1} = \left( { - \frac{{16}}{3}; - 3} \right)\\
TH2:\,\,\,\, - 3 \le x \le 4 \Rightarrow \left\{ \begin{array}{l}
4 - x > 0\\
x + 3 > 0
\end{array} \right. \Rightarrow \left\{ \begin{array}{l}
\left| {4 - x} \right| = 4 - x\\
\left| {x + 3} \right| = x + 3
\end{array} \right.\\
\left( 1 \right) \Leftrightarrow 4 - x + x + 3 - 3\left( {x + 3} \right) > 0\\
\Leftrightarrow - 2 - 3x > 0\\
\Leftrightarrow 3x + 2 < 0\\
\Leftrightarrow x < - \frac{2}{3}\\
\Rightarrow {S_2} = \left[ { - 3; - \frac{2}{3}} \right)\\
TH3:\,\,\,\,x > 4 \Rightarrow \left\{ \begin{array}{l}
\left| {4 - x} \right| = x - 4\\
\left| {x + 3} \right| = x + 3
\end{array} \right.\\
\left( 1 \right) \Leftrightarrow x - 4 + x + 3 - 3\left( {x + 3} \right) > 0\\
\Leftrightarrow - x - 10 > 0\\
\Leftrightarrow x + 10 < 0\\
\Leftrightarrow x < - 10\\
\Rightarrow {S_3} = \emptyset \\
\Rightarrow S = {S_1} \cup {S_2} \cup {S_3} = \left( { - \frac{{16}}{3}; - \frac{2}{3}} \right)
\end{array}\)
