1.
(x + 7)(x - 2) > 0
TH1: \(\left\{{}\begin{matrix}x+7>0\\x-2>0\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}x>-7\\x>2\end{matrix}\right.\) \(\Rightarrow x>2\)
TH2: \(\left\{{}\begin{matrix}x+7< 0\\x-2< 0\end{matrix}\right.\)\(\Rightarrow\left\{{}\begin{matrix}x< -7\\x< 2\end{matrix}\right.\) \(\Rightarrow x< -7\)
2.
\(\dfrac{37-x}{x+13}=\dfrac{3}{7}\) \(\Rightarrow7\left(37-x\right)=3\left(x+13\right)\)
\(\Leftrightarrow259-7x=3x+39\)
\(\Leftrightarrow259-39=3x+7x\)
\(\Leftrightarrow220=10x\Rightarrow x=22\)
3.
\(\dfrac{x-3}{x+8}< 0\)
TH1: \(\left\{{}\begin{matrix}x-3< 0\\x+8>0\end{matrix}\right.\) => \(\left\{{}\begin{matrix}x< 3\\x>-8\end{matrix}\right.\) => -8 < x < 3
TH2: \(\left\{{}\begin{matrix}x-3>0\\x+8< 0\end{matrix}\right.\) => \(\left\{{}\begin{matrix}x>3\\x< -8\end{matrix}\right.\) (loại)
Vậy -8 < x < 3
