Đáp án:
\[x = 4 + 2\sqrt 3 \]
Giải thích các bước giải:
Ta có:
\(\begin{array}{l}
P\left( {\sqrt x + 1} \right) = - \left( {\sqrt 3 + 3} \right).\sqrt x + 2\sqrt 3 + 3\\
\Leftrightarrow \left( {1 - \sqrt x } \right)\left( {\sqrt x + 1} \right) = - \left( {3 + \sqrt 3 } \right)\sqrt x + 2\sqrt 3 + 3\\
\Leftrightarrow {1^2} - {\sqrt x ^2} = - \left( {3 + \sqrt 3 } \right)\sqrt x + 2\sqrt 3 + 3\\
\Leftrightarrow 1 - x = - \left( {3 + \sqrt 3 } \right)\sqrt x + 2\sqrt 3 + 3\\
\Leftrightarrow x - \left( {3 + \sqrt 3 } \right)\sqrt x + 2\sqrt 3 + 2 = 0\\
\Leftrightarrow \left( {x - 2\sqrt x } \right) - \left( {1 + \sqrt 3 } \right)\sqrt x + \left( {2\sqrt 3 + 2} \right) = 0\\
\Leftrightarrow \sqrt x \left( {\sqrt x - 2} \right) - \left( {1 + \sqrt 3 } \right)\left( {\sqrt x - 2} \right) = 0\\
\Leftrightarrow \left( {\sqrt x - 2} \right)\left( {\sqrt x - \left( {1 + \sqrt 3 } \right)} \right) = 0\\
\Leftrightarrow \left[ \begin{array}{l}
\sqrt x - 2 = 0\\
\sqrt x - \left( {1 + \sqrt 3 } \right) = 0
\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}
\sqrt x = 2\\
\sqrt x = 1 + \sqrt 3
\end{array} \right.\\
\Leftrightarrow \left[ \begin{array}{l}
x = 4\\
x = 4 + 2\sqrt 3
\end{array} \right.\\
x > 0,\,\,x \ne 4 \Rightarrow x = 4 + 2\sqrt 3
\end{array}\)
Vậy \(x = 4 + 2\sqrt 3 \)
