Đáp án:
2) \(0 \le x < 9\)
Giải thích các bước giải:
\(\begin{array}{l}
1)Thay:x = 36\\
Q = \dfrac{{\sqrt {36} + 1}}{{\sqrt {36} - 3}} = \dfrac{{6 + 1}}{{6 - 3}} = \dfrac{7}{3}\\
2)P = \dfrac{{2\sqrt x \left( {\sqrt x - 3} \right) + \sqrt x \left( {\sqrt x + 3} \right) - 3x - 3}}{{\left( {\sqrt x - 3} \right)\left( {\sqrt x + 3} \right)}}\\
= \dfrac{{2x - 6\sqrt x + x + 3\sqrt x - 3x - 3}}{{\left( {\sqrt x - 3} \right)\left( {\sqrt x + 3} \right)}}\\
= \dfrac{{ - 3\sqrt x - 3}}{{\left( {\sqrt x - 3} \right)\left( {\sqrt x + 3} \right)}}\\
\dfrac{P}{Q} = \dfrac{{ - 3\sqrt x - 3}}{{\left( {\sqrt x - 3} \right)\left( {\sqrt x + 3} \right)}}:\dfrac{{\sqrt x + 1}}{{\sqrt x - 3}}\\
= \dfrac{{ - 3\left( {\sqrt x + 1} \right)}}{{\left( {\sqrt x - 3} \right)\left( {\sqrt x + 3} \right)}}.\dfrac{{\sqrt x - 3}}{{\sqrt x + 1}}\\
= - \dfrac{3}{{\sqrt x + 3}}\\
\dfrac{P}{Q} < - \dfrac{1}{2}\\
\to - \dfrac{3}{{\sqrt x + 3}} < - \dfrac{1}{2}\\
\to \dfrac{3}{{\sqrt x + 3}} > \dfrac{1}{2}\\
\to \dfrac{{6 - \sqrt x - 3}}{{2\left( {\sqrt x + 3} \right)}} > 0\\
\to \dfrac{{3 - \sqrt x }}{{2\left( {\sqrt x + 3} \right)}} > 0\\
\to 3 - \sqrt x > 0\left( {do:\sqrt x + 3 > 0} \right)\\
\to 9 > x\\
\to 0 \le x < 9
\end{array}\)
