Đáp án:
\(\begin{array}{l}
TH1:P > \dfrac{1}{2}\\
\Leftrightarrow x > 9\\
TH2:P < \dfrac{1}{2}\\
\Leftrightarrow 0 < x < 9
\end{array}\)
Giải thích các bước giải:
\(\begin{array}{l}
c)P = \dfrac{{\sqrt x }}{{\sqrt x - 3}}\\
Xét:P - \dfrac{1}{2} = \dfrac{{\sqrt x }}{{\sqrt x - 3}} - \dfrac{1}{2}\\
= \dfrac{{2\sqrt x - \sqrt x + 3}}{{2\left( {\sqrt x - 3} \right)}}\\
= \dfrac{{\sqrt x + 3}}{{2\left( {\sqrt x - 3} \right)}}\\
TH1:P - \dfrac{1}{2} > 0 \to P > \dfrac{1}{2}\\
\Leftrightarrow \dfrac{{\sqrt x + 3}}{{2\left( {\sqrt x - 3} \right)}} > 0\\
\Leftrightarrow \sqrt x - 3 > 0\left( {do:\sqrt x + 3 > 0\forall x > 0} \right)\\
\to x > 9\\
TH2:P < \dfrac{1}{2} > 0 \to P < \dfrac{1}{2}\\
\Leftrightarrow \dfrac{{\sqrt x + 3}}{{2\left( {\sqrt x - 3} \right)}} < 0\\
\Leftrightarrow \sqrt x - 3 < 0\left( {do:\sqrt x + 3 > 0\forall x > 0} \right)\\
\to 0 < x < 9
\end{array}\)
