а) \(\left( {x + 1} \right)\sqrt {{x^2} + 4x-8} = 2\left( {x + 1} \right).\)
Запишем ОДЗ:
\({x^2} + 4x-8 \ge 0\;\;\;\; \Leftrightarrow \;\;\;\;\left( {x + 2 + 2\sqrt 3 } \right)\left( {x + 2-2\sqrt 3 } \right) \ge 0\;\;\;\; \Leftrightarrow \;\;\;\;x \in \left( {-\infty ;-2-2\sqrt 3 } \right] \cup \left[ {-2 + 2\sqrt 3 ;\infty } \right).\)
\(\left( {x + 1} \right)\sqrt {{x^2} + 4x-8} = 2\left( {x + 1} \right)\;\;\;\; \Leftrightarrow \;\;\;\;\left( {x + 1} \right)\sqrt {{x^2} + 4x-8} -2\left( {x + 1} \right) = 0\;\;\;\; \Leftrightarrow \)
\( \Leftrightarrow \;\;\;\;\left( {x + 1} \right)\left( {\sqrt {{x^2} + 4x-8} -2} \right) = 0\;\;\;\; \Leftrightarrow \;\;\;\;\left[ {\begin{array}{*{20}{c}}{x + 1 = 0,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,}\\{\sqrt {{x^2} + 4x-8} = 2}\end{array}\;\;\;\; \Leftrightarrow } \right.\)
\( \Leftrightarrow \;\;\;\;\left[ {\begin{array}{*{20}{c}}{x = -1 \notin \left( {-\infty ;-2-2\sqrt 3 } \right] \cup \left[ {-2 + 2\sqrt 3 ;\infty } \right),}\\{{x^2} + 4x-8 = 4\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,}\end{array}} \right.\;\;\;\; \Leftrightarrow \;\;\;\;{x^2} + 4x-12 = 0\;\;\;\; \Leftrightarrow \;\;\;\;\left[ {\begin{array}{*{20}{c}}{{x} = 2,\;\;}\\{{x} = -6.}\end{array}} \right.\)
б) Отберём корни, принадлежащие отрезку \(\left[ {-\sqrt {35} ;\,\,\sqrt 5 } \right].\)
Так как \(-\sqrt {35} < 2 = \sqrt 4 < \sqrt 5 ,\) то \(x = 2 \in \left[ {-\sqrt {35} ;\sqrt 5 } \right].\)
Так как \(-6 = -\sqrt {36} < -\sqrt {35} ,\) то \(x = -6 \notin \left[ {-\sqrt 3 5;\sqrt 5 } \right].\)
Ответ: а) \(-6;\;\;\;\;2;\)
б) \(2.\)