Đáp án:
\[4 < x < 36\]
Giải thích các bước giải:
Ta có:
\(\begin{array}{l}
A > \dfrac{5}{4}\\
\Leftrightarrow \dfrac{{\sqrt x - 1}}{{\sqrt x - 2}} > \dfrac{5}{4}\\
\Leftrightarrow \dfrac{{\sqrt x - 1}}{{\sqrt x - 2}} - \dfrac{5}{4} > 0\\
\Leftrightarrow \dfrac{{4.\left( {\sqrt x - 1} \right) - 5.\left( {\sqrt x - 2} \right)}}{{4.\left( {\sqrt x - 2} \right)}} > 0\\
\Leftrightarrow \dfrac{{\left( {4\sqrt x - 4} \right) - \left( {5\sqrt x - 10} \right)}}{{4\left( {\sqrt x - 2} \right)}} > 0\\
\Leftrightarrow \dfrac{{4\sqrt x - 4 - 5\sqrt x + 10}}{{4\left( {\sqrt x - 2} \right)}} > 0\\
\Leftrightarrow \dfrac{{6 - \sqrt x }}{{4.\left( {\sqrt x - 2} \right)}} > 0\\
\Leftrightarrow \dfrac{{6 - \sqrt x }}{{\sqrt x - 2}} > 0\\
\Leftrightarrow \left[ \begin{array}{l}
\left\{ \begin{array}{l}
6 - \sqrt x > 0\\
\sqrt x - 2 > 0
\end{array} \right.\\
\left\{ \begin{array}{l}
6 - \sqrt x < 0\\
\sqrt x - 2 < 0
\end{array} \right.
\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}
\left\{ \begin{array}{l}
\sqrt x < 6\\
\sqrt x > 2
\end{array} \right.\\
\left\{ \begin{array}{l}
\sqrt x > 6\\
\sqrt x < 2
\end{array} \right.
\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}
2 < \sqrt x < 6\\
6 < \sqrt x < 2\,\,\,\left( L \right)
\end{array} \right.\\
\Leftrightarrow 2 < \sqrt x < 6\\
\Leftrightarrow 4 < x < 36
\end{array}\)
Vậy \(4 < x < 36\)
