Đáp án:
$\begin{array}{l}
a)Dkxd:x \ge 0\\
M = x - 4\\
\Rightarrow \dfrac{{x\sqrt x - 8}}{{3 + {{\left( {\sqrt x + 1} \right)}^2}}} = x - 4\\
\Rightarrow \dfrac{{\left( {\sqrt x - 2} \right)\left( {x + 2\sqrt x + 4} \right)}}{{3 + x + 2\sqrt x + 1}} = \left( {\sqrt x - 2} \right)\left( {\sqrt x + 2} \right)\\
\Rightarrow \left( {\sqrt x - 2} \right).\left( {\dfrac{{x + 2\sqrt x + 4}}{{x + 2\sqrt x + 4}} - \sqrt x - 2} \right) = 0\\
\Rightarrow \left[ \begin{array}{l}
\sqrt x - 2 = 0\\
1 - \sqrt x - 2 = 0
\end{array} \right.\\
\Rightarrow \left[ \begin{array}{l}
\sqrt x = 2\\
\sqrt x = - 1\left( {ktm} \right)
\end{array} \right.\\
\Rightarrow x = 4\left( {tm} \right)\\
Vậy\,x = 4\\
b)M = \dfrac{{\left( {\sqrt x - 2} \right)\left( {x + 2\sqrt x + 4} \right)}}{{x + 2\sqrt x + 4}}\\
= \sqrt x - 2\\
N = \dfrac{{{{\left( {\sqrt x + 1} \right)}^3} - {{\left( {\sqrt x - 1} \right)}^3}}}{{\left( {x - 4} \right)\left( {3x + 1} \right)}}\\
= \dfrac{{\left( {\sqrt x + 1 - \sqrt x + 1} \right)\left[ {{{\left( {\sqrt x + 1} \right)}^2} + \left( {\sqrt x + 1} \right)\left( {\sqrt x - 1} \right) + {{\left( {\sqrt x - 1} \right)}^2}} \right]}}{{\left( {x - 4} \right)\left( {3x + 1} \right)}}\\
= \dfrac{{2.\left( {x + 2\sqrt x + 1 + x - 1 + x - 2\sqrt x + 1} \right)}}{{\left( {x - 4} \right)\left( {3x + 1} \right)}}\\
= \dfrac{{2.\left( {3x + 1} \right)}}{{\left( {x - 4} \right)\left( {3x + 1} \right)}}\\
= \dfrac{2}{{x - 4}}\\
\Rightarrow Q = M.N + P\\
= \left( {\sqrt x - 2} \right).\dfrac{2}{{x - 4}} + \dfrac{{\sqrt x }}{{2 + \sqrt x }}\\
= \dfrac{2}{{\sqrt x + 2}} + \dfrac{{\sqrt x }}{{\sqrt x + 2}}\\
= \dfrac{{2 + \sqrt x }}{{\sqrt x + 2}}\\
= 1
\end{array}$
