Tìm x, biết: a, $\frac{x-2}{x-6}$ < 0 b, $x^{2}$ + 4x > 0

Đáp án:

$\begin{array}{l}
a)\dfrac{{x – 2}}{{x – 6}} < 0\\
 \Rightarrow \left[ \begin{array}{l}
\left\{ \begin{array}{l}
x – 2 > 0\\
x – 6 < 0
\end{array} \right.\\
\left\{ \begin{array}{l}
x – 2 < 0\\
x – 6 > 0
\end{array} \right.
\end{array} \right. \Rightarrow \left[ \begin{array}{l}
\left\{ \begin{array}{l}
x > 2\\
x < 6
\end{array} \right.\\
\left\{ \begin{array}{l}
x < 2\\
x > 6
\end{array} \right.\left( {ktm} \right)
\end{array} \right.\\
 \Rightarrow 2 < x < 6\\
\text{Vậy}\,2 < x < 6\\
b){x^2} + 4x > 0\\
 \Rightarrow x\left( {x + 4} \right) > 0\\
 \Rightarrow \left[ \begin{array}{l}
\left\{ \begin{array}{l}
x > 0\\
x + 4 > 0
\end{array} \right.\\
\left\{ \begin{array}{l}
x < 0\\
x + 4 < 0
\end{array} \right.
\end{array} \right. \Rightarrow \left[ \begin{array}{l}
\left\{ \begin{array}{l}
x > 0\\
x >  – 4
\end{array} \right.\\
\left\{ \begin{array}{l}
x < 0\\
x <  – 4
\end{array} \right.
\end{array} \right.\\
 \Rightarrow \left[ \begin{array}{l}
x > 0\\
x <  – 4
\end{array} \right.\\
\text{Vậy}\,x > 0\,\text{hoặc}\,x <  – 4
\end{array}$