Đáp án:
c. \(\left[ \begin{array}{l}
x > \dfrac{2}{3}\\
x < - 4
\end{array} \right.\)
Giải thích các bước giải:
\(\begin{array}{l}
a.\dfrac{{2x + 1}}{6} - x > 3 - \dfrac{{x - 2}}{9}\\
\to \dfrac{{9\left( {2x + 1} \right) - 36x}}{{36}} > \dfrac{{3.36 - 6\left( {x - 2} \right)}}{{36}}\\
\to 9\left( {2x + 1} \right) - 36x > 3.36 - 6\left( {x - 2} \right)\\
\to 18x + 9 - 36x > 108 - 6x + 12\\
\to 12x < - 111\\
\to x < - \dfrac{{111}}{{12}}\\
b.{x^2} - 2x < 0\\
\to x\left( {x - 2} \right) < 0\\
\to \left[ \begin{array}{l}
\left\{ \begin{array}{l}
x > 0\\
x - 2 < 0
\end{array} \right.\\
\left\{ \begin{array}{l}
x < 0\\
x - 2 > 0
\end{array} \right.
\end{array} \right. \to \left[ \begin{array}{l}
\left\{ \begin{array}{l}
x > 0\\
x < 2
\end{array} \right.\\
\left\{ \begin{array}{l}
x < 0\\
x > 2
\end{array} \right.\left( l \right)
\end{array} \right.\\
c.\dfrac{{3x - 2}}{{x + 4}} > 0\\
\to \left[ \begin{array}{l}
\left\{ \begin{array}{l}
3x - 2 > 0\\
x + 4 > 0
\end{array} \right.\\
\left\{ \begin{array}{l}
3x - 2 < 0\\
x + 4 < 0
\end{array} \right.
\end{array} \right. \to \left[ \begin{array}{l}
\left\{ \begin{array}{l}
x > \dfrac{2}{3}\\
x > - 4
\end{array} \right.\\
\left\{ \begin{array}{l}
x < \dfrac{2}{3}\\
x < - 4
\end{array} \right.
\end{array} \right.\\
\to \left[ \begin{array}{l}
x > \dfrac{2}{3}\\
x < - 4
\end{array} \right.
\end{array}\)
