2 câu trả lời
Đáp án:
\(16 > x > 4\)
Giải thích các bước giải:
\(\begin{array}{l}
\dfrac{{3\sqrt x }}{{\sqrt x - 2}} > \sqrt x + 2\\
\to \dfrac{{3\sqrt x - \left( {\sqrt x + 2} \right)\left( {\sqrt x - 2} \right)}}{{\sqrt x - 2}} > 0\\
\to \dfrac{{3\sqrt x - x + 4}}{{\sqrt x - 2}} > 0\\
\to \dfrac{{\left( {4 - \sqrt x } \right)\left( {\sqrt x + 1} \right)}}{{\sqrt x - 2}} > 0\\
\to \dfrac{{4 - \sqrt x }}{{\sqrt x - 2}} > 0\left( {do:\sqrt x + 1 > 0} \right)\\
\to \left[ \begin{array}{l}
\left\{ \begin{array}{l}
4 - \sqrt x > 0\\
\sqrt x - 2 > 0
\end{array} \right.\\
\left\{ \begin{array}{l}
4 - \sqrt x < 0\\
\sqrt x - 2 < 0
\end{array} \right.
\end{array} \right.\\
\to \left[ \begin{array}{l}
4 > \sqrt x > 2\\
\left\{ \begin{array}{l}
4 < \sqrt x \\
\sqrt x < 2
\end{array} \right.\left( l \right)
\end{array} \right.\\
\to 16 > x > 4
\end{array}\)