Đáp án:
$\begin{array}{l}
Dkxd:a \ge 0;a \ne 4\\
P = \frac{{\sqrt a + 2}}{{\sqrt a + 3}} - \frac{5}{{a + \sqrt a - 6}} + \frac{1}{{2 - \sqrt a }}\\
= \frac{{\left( {\sqrt a + 2} \right).\left( {\sqrt a - 2} \right) - 5 - \left( {\sqrt a + 3} \right)}}{{\left( {\sqrt a + 3} \right)\left( {\sqrt a - 2} \right)}}\\
= \frac{{a - 4 - 5 - \sqrt a - 3}}{{\left( {\sqrt a + 3} \right)\left( {\sqrt a - 2} \right)}}\\
= \frac{{a - \sqrt a - 12}}{{\left( {\sqrt a + 3} \right)\left( {\sqrt a - 2} \right)}}\\
= \frac{{\left( {\sqrt a + 3} \right)\left( {\sqrt a - 4} \right)}}{{\left( {\sqrt a + 3} \right)\left( {\sqrt a - 2} \right)}}\\
= \frac{{\sqrt a - 4}}{{\sqrt a - 2}}\\
P < 1\\
\Rightarrow \frac{{\sqrt a - 4}}{{\sqrt a - 2}} < 1\\
\Rightarrow \frac{{\sqrt a - 4 - \left( {\sqrt a - 2} \right)}}{{\sqrt a - 2}} < 0\\
\Rightarrow \frac{{ - 2}}{{\sqrt a - 2}} < 0\\
\Rightarrow \sqrt a - 2 > 0\\
\Rightarrow \sqrt a > 2\\
\Rightarrow a > 4\\
Vậy\,a > 4
\end{array}$
