Đáp án:

\(\left[ \begin{array}{l}
x = \dfrac{{5 - \sqrt 6  + \sqrt {31 - 14\sqrt 6 } }}{2}\\
x = \dfrac{{5 - \sqrt 6  - \sqrt {31 - 14\sqrt 6 } }}{2}\\
x = \dfrac{{\sqrt 6  + 5 + \sqrt {31 + 14\sqrt 6 } }}{2}\\
x = \dfrac{{\sqrt 6  + 5 - \sqrt {31 + 14\sqrt 6 } }}{2}
\end{array} \right.\)

Giải thích các bước giải:

\(\begin{array}{l}
DK:x \ne  - 1\\
x.\left( {\dfrac{{5 - x}}{{x + 1}}} \right).x.\left( {\dfrac{{5 - x}}{{x + 1}}} \right) = 6\\
 \to {\left[ {x.\left( {\dfrac{{5 - x}}{{x + 1}}} \right)} \right]^2} = 6\\
 \to \left[ \begin{array}{l}
x.\left( {\dfrac{{5 - x}}{{x + 1}}} \right) = \sqrt 6 \\
x.\left( {\dfrac{{5 - x}}{{x + 1}}} \right) =  - \sqrt 6 
\end{array} \right.\\
 \to \left[ \begin{array}{l}
5x - {x^2} = x\sqrt 6  + \sqrt 6 \\
5x - {x^2} =  - x\sqrt 6  - \sqrt 6 
\end{array} \right.\\
 \to \left[ \begin{array}{l}
{x^2} + \left( {\sqrt 6  - 5} \right)x + \sqrt 6  = 0\left( 1 \right)\\
{x^2} - \left( {\sqrt 6  + 5} \right) - \sqrt 6  = 0\left( 2 \right)
\end{array} \right.\\
Xét\left( 1 \right): Δ= 31 - 14\sqrt 6 \\
 \to \left[ \begin{array}{l}
x = \dfrac{{5 - \sqrt 6  + \sqrt {31 - 14\sqrt 6 } }}{2}\\
x = \dfrac{{5 - \sqrt 6  - \sqrt {31 - 14\sqrt 6 } }}{2}
\end{array} \right.\\
Xét:\left( 2 \right): Δ= 31 + 14\sqrt 6 \\
 \to \left[ \begin{array}{l}
x = \dfrac{{\sqrt 6  + 5 + \sqrt {31 + 14\sqrt 6 } }}{2}\\
x = \dfrac{{\sqrt 6  + 5 - \sqrt {31 + 14\sqrt 6 } }}{2}
\end{array} \right.\\
KL:\left[ \begin{array}{l}
x = \dfrac{{5 - \sqrt 6  + \sqrt {31 - 14\sqrt 6 } }}{2}\\
x = \dfrac{{5 - \sqrt 6  - \sqrt {31 - 14\sqrt 6 } }}{2}\\
x = \dfrac{{\sqrt 6  + 5 + \sqrt {31 + 14\sqrt 6 } }}{2}\\
x = \dfrac{{\sqrt 6  + 5 - \sqrt {31 + 14\sqrt 6 } }}{2}
\end{array} \right.
\end{array}\)