Điều kiện xác định $x\ge 1$
$\begin{array}{l} \sqrt {{x^2} - 1} - 6 = 3\sqrt {x + 1} - 2\sqrt {x - 1} \\ \Leftrightarrow \sqrt {\left( {x - 1} \right)\left( {x + 1} \right)} - 6 + 2\sqrt {x - 1} - 3\sqrt {x + 1} = 0\\ \Leftrightarrow \sqrt {x - 1} \left( {\sqrt {x + 1} + 2} \right) - 3\left( {\sqrt {x + 1} + 2} \right) = 0\\ \Leftrightarrow \left( {\sqrt {x + 1} + 2} \right)\left( {\sqrt {x - 1} - 3} \right) = 0\\ \Leftrightarrow \left[ \begin{array}{l} \sqrt {x + 1} = - 2(L\,do \,{\sqrt{x+1}\ge 0})\\ \sqrt {x - 1} = 3 \end{array} \right.\\ \Rightarrow \sqrt {x - 1} = 3 \Leftrightarrow x - 1 = 9 \Leftrightarrow x = 10(TM)\\ \Rightarrow S = \left\{ {10} \right\} \end{array}$
