Đáp án:
\(\begin{array}{l}
a,\\
P = \dfrac{{ - 3}}{{\sqrt x + 3}}\\
b,\\
\left\{ \begin{array}{l}
0 \le x < 36\\
x \ne 9
\end{array} \right.
\end{array}\)
Giải thích các bước giải:
Ta có:
\(\begin{array}{l}
a,\\
P = \left( {\dfrac{{2\sqrt x }}{{\sqrt x + 3}} + \dfrac{{\sqrt x }}{{\sqrt x - 3}} + \dfrac{{3x + 3}}{{9 - x}}} \right):\dfrac{{\sqrt x + 1}}{{\sqrt x - 3}}\\
= \left( {\dfrac{{2\sqrt x }}{{\sqrt x + 3}} + \dfrac{{\sqrt x }}{{\sqrt x - 3}} - \dfrac{{3x + 3}}{{x - 9}}} \right):\dfrac{{\sqrt x + 1}}{{\sqrt x - 3}}\\
= \left( {\dfrac{{2\sqrt x }}{{\sqrt x + 3}} + \dfrac{{\sqrt x }}{{\sqrt x - 3}} - \dfrac{{3x + 3}}{{{{\sqrt x }^2} - {3^2}}}} \right):\dfrac{{\sqrt x + 1}}{{\sqrt x - 3}}\\
= \left( {\dfrac{{2\sqrt x }}{{\sqrt x + 3}} + \dfrac{{\sqrt x }}{{\sqrt x - 3}} - \dfrac{{3x + 3}}{{\left( {\sqrt x - 3} \right)\left( {\sqrt x + 3} \right)}}} \right):\dfrac{{\sqrt x + 1}}{{\sqrt x - 3}}\\
= \dfrac{{2\sqrt x \left( {\sqrt x - 3} \right) + \sqrt x \left( {\sqrt x + 3} \right) - \left( {3x + 3} \right)}}{{\left( {\sqrt x - 3} \right)\left( {\sqrt x + 3} \right)}}:\dfrac{{\sqrt x + 1}}{{\sqrt x - 3}}\\
= \dfrac{{2x - 6\sqrt x + x + 3\sqrt x - 3x - 3}}{{\left( {\sqrt x - 3} \right)\left( {\sqrt x + 3} \right)}}:\dfrac{{\sqrt x + 1}}{{\sqrt x - 3}}\\
= \dfrac{{ - 3\sqrt x - 3}}{{\left( {\sqrt x - 3} \right)\left( {\sqrt x + 3} \right)}}.\dfrac{{\sqrt x - 3}}{{\sqrt x + 1}}\\
= \dfrac{{ - 3.\left( {\sqrt x + 1} \right)}}{{\left( {\sqrt x - 3} \right)\left( {\sqrt x + 3} \right)}}.\dfrac{{\sqrt x - 3}}{{\sqrt x + 1}}\\
= \dfrac{{ - 3}}{{\sqrt x + 3}}\\
b,\\
P < - \dfrac{1}{3}\\
\Leftrightarrow \dfrac{{ - 3}}{{\sqrt x + 3}} + \dfrac{1}{3} < 0\\
\Leftrightarrow \dfrac{{ - 9 + \left( {\sqrt x + 3} \right)}}{{3.\left( {\sqrt x + 3} \right)}} < 0\\
\Leftrightarrow \dfrac{{\sqrt x - 6}}{{3\left( {\sqrt x + 3} \right)}} < 0\\
\sqrt x + 3 \ge 3 > 0,\,\,\,\forall x \ge 0,x \ne 9\\
\Rightarrow \sqrt x - 6 < 0\\
\Leftrightarrow \sqrt x < 6\\
\Leftrightarrow x < 36\\
\Rightarrow \left\{ \begin{array}{l}
0 \le x < 36\\
x \ne 9
\end{array} \right.
\end{array}\)
