\(\dfrac{x}{{\sqrt {4-x} + 2}} + \dfrac{x}{{\sqrt {4 + x} -2}} = 2\)
Запишем ОДЗ: \(\left\{ {\begin{array}{*{20}{c}}{x \ge -4,}\\{x \le 4,}\\{x \ne 0\,\,\,}\end{array}} \right.\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,x\, \in \,\left[ {-4;0} \right) \cup \left( {0;4} \right].\)
\(\dfrac{{x\left( {\sqrt {4-x} -2} \right)}}{{\left( {\sqrt {4-x} + 2} \right)\left( {\sqrt {4-x} -2} \right)}} + \dfrac{{x\left( {\sqrt {4 + x} + 2} \right)}}{{\left( {\sqrt {4 + x} -2} \right)\left( {\sqrt {4 + x} + 2} \right)}} = 2\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\dfrac{{x\left( {\sqrt {4-x} -2} \right)}}{{4-x-4}} + \dfrac{{x\left( {\sqrt {4 + x} + 2} \right)}}{{4 + x-4}} = 2\,\,\,\,\,\, \Leftrightarrow \)
\( \Leftrightarrow \,\,\,\,\,2-\sqrt {4-x} + \sqrt {4 + x} + 2 = 2\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\,2 + \sqrt {4 + x} = \sqrt {4-x} \,\,\,\,\,\, \Leftrightarrow \)
\( \Leftrightarrow \,\,\,\,\,4 + 4\sqrt {4 + x} + 4 + x = 4-x\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,4\sqrt {4 + x} = -2x-4\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,2\sqrt {4 + x} = -x-2\,\,\,\,\,\, \Leftrightarrow \)
\( \Leftrightarrow \,\,\,\,\,\left\{ {\begin{array}{*{20}{c}}{-x-2 \ge 0,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,}\\{16 + 4x = {x^2} + 4x + 4}\end{array}} \right.\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\left\{ {\begin{array}{*{20}{c}}{x \le -2,\,\,\,\,\,\,\,}\\{\left[ {\begin{array}{*{20}{c}}{x = 2\sqrt 3 ,\,\,}\\{x = -2\sqrt 3 }\end{array}} \right.}\end{array}} \right.\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,x = -2\sqrt 3 .\)
Ответ: \(-2\sqrt 3 \).