Đáp án:
$\begin{array}{l}
Dkxd:x \ge 0\\
P.\left( {\sqrt x + 1} \right) = - \left( {\sqrt 3 + 3} \right).\sqrt x + 2\sqrt 3 + 3\\
\Rightarrow \left( {1 - \sqrt x } \right).\left( {\sqrt x + 1} \right)\\
= - \left( {\sqrt 3 + 3} \right).\sqrt x + 2\sqrt 3 + 3\\
\Rightarrow 1 - x = - \left( {\sqrt 3 + 3} \right).\sqrt x + 2\sqrt 3 + 3\\
\Rightarrow x - \left( {\sqrt 3 + 3} \right).\sqrt x + 2\sqrt 3 + 2 = 0\\
\Rightarrow x - 2\sqrt x - \left( {\sqrt 3 + 1} \right).\sqrt x + 2\left( {\sqrt 3 + 1} \right) = 0\\
\Rightarrow \left( {\sqrt x - 2} \right).\left( {\sqrt x - \sqrt 3 - 1} \right) = 0\\
\Rightarrow \left[ \begin{array}{l}
\sqrt x = 2\\
\sqrt x = \sqrt 3 - 1
\end{array} \right.\\
\Rightarrow \left[ \begin{array}{l}
x = 4\left( {tm} \right)\\
x = {\left( {\sqrt 3 - 1} \right)^2} = 4 - 2\sqrt 3 \left( {tm} \right)
\end{array} \right.
\end{array}$
