Đáp án:
\(0 < a < 16;a \ne \left\{ {4;9} \right\}\)
Giải thích các bước giải:
\(\begin{array}{l}
a)DK:a > 0;a \ne \left\{ {4;9} \right\}\\
P = \dfrac{{9\sqrt a - 4a + \left( {2\sqrt a + 3} \right)\left( {\sqrt a - 2} \right) + \left( {\sqrt a + 1} \right)\left( {\sqrt a - 3} \right)}}{{\left( {\sqrt a - 2} \right)\left( {\sqrt a - 3} \right)}}.\dfrac{{\sqrt a \left( {\sqrt a - 2} \right)\left( {\sqrt a + 2} \right)}}{{\sqrt a + 2}}\\
= \dfrac{{9\sqrt a - 4a + 2a - \sqrt a - 6 + a - 2\sqrt a - 3}}{{\left( {\sqrt a - 2} \right)\left( {\sqrt a - 3} \right)}}.\dfrac{{\sqrt a \left( {\sqrt a - 2} \right)\left( {\sqrt a + 2} \right)}}{{\sqrt a + 2}}\\
= \dfrac{{ - a + 6\sqrt a - 9}}{{\left( {\sqrt a - 2} \right)\left( {\sqrt a - 3} \right)}}.\sqrt a \left( {\sqrt a - 2} \right)\\
= \dfrac{{ - {{\left( {\sqrt a - 3} \right)}^2}}}{{\left( {\sqrt a - 2} \right)\left( {\sqrt a - 3} \right)}}.\sqrt a \left( {\sqrt a - 2} \right)\\
= - \sqrt a \left( {\sqrt a - 3} \right)\\
= - a + 3\sqrt a \\
\left| P \right| > 4\\
\to \left[ \begin{array}{l}
P > 4\\
P < - 4
\end{array} \right.\\
\to \left[ \begin{array}{l}
- a + 3\sqrt a > 4\\
- a + 3\sqrt a < - 4
\end{array} \right.\\
\to \left[ \begin{array}{l}
- a + 3\sqrt a - 4 > 0\\
- a + 3\sqrt a + 4 < 0
\end{array} \right.\\
\to \left[ \begin{array}{l}
a - 3\sqrt a + 4 > 0\\
a - 3\sqrt a - 4 < 0
\end{array} \right.\\
\to \left[ \begin{array}{l}
a - 2\sqrt a .\dfrac{3}{2} + \dfrac{9}{4} + \dfrac{7}{4} > 0\\
\left( {\sqrt a - 4} \right)\left( {\sqrt a + 1} \right) < 0
\end{array} \right.\\
\to \left[ \begin{array}{l}
{\left( {\sqrt a + \dfrac{3}{2}} \right)^2} + \dfrac{7}{4} > 0\\
\sqrt a - 4 < 0\left( {do:\sqrt a + 1 > 0\forall a > 0} \right)
\end{array} \right.\\
\to 0 < a < 16;a \ne \left\{ {4;9} \right\}
\end{array}\)
