\(\left( {x + 1} \right)\sqrt {{x^2} + 4x-8} = 2\left( {x + 1} \right).\)
Запишем ОДЗ:
\({x^2} + 4x-8 \ge 0\;\;\;\; \Leftrightarrow \;\;\;\;\left( {x + 2 + 2\sqrt 3 } \right)\left( {x + 2-2\sqrt 3 } \right) \ge 0\;\;\;\; \Leftrightarrow \;\;\;\;x \in \left( {-\infty ;-2-2\sqrt 3 } \right] \cup \left[ {-2 + 2\sqrt 3 ;\infty } \right).\)
\(\left( {x + 1} \right)\sqrt {{x^2} + 4x-8} = 2\left( {x + 1} \right)\;\;\;\; \Leftrightarrow \;\;\;\;\left( {x + 1} \right)\sqrt {{x^2} + 4x-8} -2\left( {x + 1} \right) = 0\;\;\;\; \Leftrightarrow \)
\( \Leftrightarrow \;\;\;\;\left( {x + 1} \right)\left( {\sqrt {{x^2} + 4x-8} -2} \right) = 0\;\;\;\; \Leftrightarrow \;\;\;\;\left[ {\begin{array}{*{20}{c}}{x + 1 = 0,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,}\\{\sqrt {{x^2} + 4x-8} = 2}\end{array}\;\;\;\; \Leftrightarrow } \right.\)
\( \Leftrightarrow \;\;\;\;\left[ {\begin{array}{*{20}{c}}{x = -1 \notin \left( {-\infty ;-2-2\sqrt 3 } \right] \cup \left[ {-2 + 2\sqrt 3 ;\infty } \right),}\\{{x^2} + 4x-8 = 4\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,}\end{array}} \right.\;\;\;\; \Leftrightarrow \;\;\;\;{x^2} + 4x-12 = 0\;\;\;\; \Leftrightarrow \;\;\;\;\left[ {\begin{array}{*{20}{c}}{{x} = 2,\;\;}\\{{x} = -6.}\end{array}} \right.\)
ОТВЕТ: -6; 2.