Đáp án:
$\begin{array}{l}
2)Q = \left( {\dfrac{{3\sqrt x }}{{\sqrt x + 2}} - \dfrac{{6x}}{{x - 4}}} \right):\dfrac{{\sqrt x }}{{2 - \sqrt x }}\\
= \dfrac{{3\sqrt x \left( {\sqrt x - 2} \right) - 6x}}{{\left( {\sqrt x + 2} \right)\left( {\sqrt x - 2} \right)}}.\dfrac{{\sqrt x - 2}}{{ - \sqrt x }}\\
= \dfrac{{3x - 6\sqrt x - 6x}}{{\sqrt x + 2}}.\dfrac{1}{{ - \sqrt x }}\\
= \dfrac{{ - 3x - 6\sqrt x }}{{\sqrt x + 2}}.\dfrac{1}{{ - \sqrt x }}\\
= \dfrac{{ - 3\sqrt x \left( {\sqrt x + 2} \right)}}{{ - \sqrt x \left( {\sqrt x + 2} \right)}}\\
= 3\\
Vậy\,Q = 3\\
3)P = Q\\
\Leftrightarrow \dfrac{{x + 1}}{{\sqrt {x + 2} - 1}} = 3\\
\Leftrightarrow x + 1 = 3\sqrt {x + 2} - 3\\
\Leftrightarrow x + 4 - 3\sqrt {x + 2} = 0\\
\Leftrightarrow x + 2 - 3\sqrt {x + 2} + 2 = 0\\
\Leftrightarrow {\left( {\sqrt {x + 2} } \right)^2} - 3\sqrt {x + 2} + 2 = 0\\
\Leftrightarrow \left( {\sqrt {x + 2} - 1} \right)\left( {\sqrt {x + 2} - 2} \right) = 0\\
\Leftrightarrow \left[ \begin{array}{l}
\sqrt {x + 2} = 1\\
\sqrt {x + 2} = 2
\end{array} \right.\\
\Leftrightarrow \left[ \begin{array}{l}
x + 2 = 1\\
x + 2 = 4
\end{array} \right.\\
\Leftrightarrow \left[ \begin{array}{l}
x = - 1\left( {ktm} \right)\\
x = 2\left( {tm} \right)
\end{array} \right.\\
Vậy\,x = 2
\end{array}$
