a) Ta biến đổi :
\(\displaystyle\eqalign{ & {{2x - 1} \over {x + 3}} > 1 \cr & \Leftrightarrow {{2x - 1} \over {x + 3}} - 1 > 0 \cr & \Leftrightarrow {{2x - 1 - \left( {x + 3} \right)} \over {x + 3}} > 0 \cr & \Leftrightarrow {{x - 4} \over {x + 3}} > 0 \cr} \)
\(\Leftrightarrow \left[ \begin{array}{l}
\left\{ \begin{array}{l}
x - 4 > 0\\
x + 3 > 0
\end{array} \right.\\
\left\{ \begin{array}{l}
x - 4 < 0\\
x + 3 < 0
\end{array} \right.
\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}
\left\{ \begin{array}{l}
x > 4\\
x > - 3
\end{array} \right.\\
\left\{ \begin{array}{l}
x < 4\\
x < - 3
\end{array} \right.
\end{array} \right. \\\Leftrightarrow \left[ \begin{array}{l}
x > 4\\
x < - 3
\end{array} \right.\)
Vậy với \(x > 4\) hoặc \(x < -3\) thì
\(\displaystyle{{2x - 1} \over {x + 3}} > 1.\)
b) Ta biến đổi:
\(\displaystyle\eqalign{ & {{2x - 1} \over {x - 2}} < 3 \cr & \Leftrightarrow {{2x - 1} \over {x - 2}} - 3 < 0 \cr & \Leftrightarrow {{2x - 1 - 3\left( {x - 2} \right)} \over {x - 2}} < 0 \cr & \Leftrightarrow {{ - x + 5} \over {x - 2}} < 0 \Leftrightarrow {{x - 5} \over {x - 2}} > 0 \cr} \)
\( \Leftrightarrow \left[ \begin{array}{l}
\left\{ \begin{array}{l}
x - 5 > 0\\
x - 2 > 0
\end{array} \right.\\
\left\{ \begin{array}{l}
x - 5 < 0\\
x - 2 < 0
\end{array} \right.
\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}
\left\{ \begin{array}{l}
x > 5\\
x > 2
\end{array} \right.\\
\left\{ \begin{array}{l}
x < 5\\
x < 2
\end{array} \right.
\end{array} \right.\\ \Leftrightarrow \left[ \begin{array}{l}
x > 5\\
x < 2
\end{array} \right.\)
Vậy với \(x > 5\) hoặc \(x < 2\) thì
\(\displaystyle{{2x - 1} \over {x - 2}} < 3.\)
