x^2-6(x+3) .căn(x+1)+14x+3 .căn (x+1)+13=0

Điều kiện xác định $x\ge -1$

$\begin{array}{l}
{x^2} - 6\left( {x + 3} \right)\sqrt {x + 1}  + 14x + 3\sqrt {x + 1}  + 13 = 0\\
 \Leftrightarrow {x^2} - \left( {6x + 15} \right)\sqrt {x + 1}  + 14x + 13 = 0\\
 \Leftrightarrow {x^2} + 14x + 13 = \left( {6x + 15} \right)\sqrt {x + 1} \\
 \Leftrightarrow {x^2} + 14x + 13 = \left( {6x + 15} \right)\left( {\sqrt {x + 1}  - 3} \right) + 3\left( {6x + 15} \right)\\
 \Leftrightarrow {x^2} - 4x - 32 = \left( {6x + 15} \right)\left( {\sqrt {x + 1}  - 3} \right)\\
 \Leftrightarrow \left( {x - 8} \right)\left( {x + 4} \right) = \left( {6x + 15} \right).\dfrac{{x - 8}}{{\sqrt {x + 1}  + 3}}\\
 \Leftrightarrow \left( {x - 8} \right)\left( {x + 4 - \dfrac{{6x + 15}}{{\sqrt {x + 1}  + 3}}} \right) = 0\\
 \Leftrightarrow \left[ \begin{array}{l}
x = 8\\
x + 4 - \dfrac{{6x + 15}}{{\sqrt {x + 1}  + 3}} = 0
\end{array} \right.\\
 \Leftrightarrow \left[ \begin{array}{l}
x = 8\\
\left( {x + 4} \right)\left( {\sqrt {x + 1}  + 3} \right) = 6x + 15
\end{array} \right.\\
 \Leftrightarrow \left[ \begin{array}{l}
x = 8\\
\left( {x + 1} \right)\sqrt {x + 1}  + 3\left( {x + 4} \right) = 6x + 15
\end{array} \right.\\
 \Leftrightarrow \left[ \begin{array}{l}
x = 8\\
\left( {x + 1} \right)\sqrt {x + 1}  = 3x + 3
\end{array} \right.\\
 \Leftrightarrow \left[ \begin{array}{l}
x = 8\\
\left( {x + 1} \right)\sqrt {x + 1}  = 3\left( {x + 1} \right)
\end{array} \right.\\
 \Leftrightarrow \left[ \begin{array}{l}
x = 8\\
\left( {x + 1} \right)\left( {\sqrt {x + 1}  - 3} \right) = 0
\end{array} \right.\\
 \Leftrightarrow \left[ \begin{array}{l}
x = 8\\
x =  - 1\\
\sqrt {x + 1}  = 3
\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}
x = 8\\
x =  - 1
\end{array} \right.\\
 \Rightarrow S = \left\{ {8; - 1} \right\}
\end{array}$