\(\dfrac{{{x^2}}}{{\sqrt {1 + {x^2}} + 1}} = \sqrt {1-{x^2}} .\)
Запишем ОДЗ: \(1-{x^2} \ge 0\,\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\,\,x\, \in \,\left[ {-1;1} \right].\)
\(\dfrac{{{x^2}}}{{\sqrt {1 + {x^2}} + 1}} = \sqrt {1-{x^2}} \,\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\,\,\dfrac{{{x^2}\left( {\sqrt {1 + {x^2}} -1} \right)}}{{\left( {\sqrt {1 + {x^2}} -1} \right)\left( {\sqrt {1 + {x^2}} + 1} \right)}} = \sqrt {1-{x^2}} \,\,\,\,\,\,\, \Leftrightarrow \)
\( \Leftrightarrow \,\,\,\,\,\,\,\dfrac{{{x^2}\left( {\sqrt {1 + {x^2}} -1} \right)}}{{1 + {x^2}-1}} = \sqrt {1-{x^2}} \,\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\,\,\sqrt {1 + {x^2}} -1 = \sqrt {1-{x^2}} \,\,\,\,\,\,\, \Leftrightarrow \)
\( \Leftrightarrow \,\,\,\,\,\,\,\sqrt {1 + {x^2}} = 1 + \sqrt {1-{x^2}} \,\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\,\,1 + {x^2} = 1 + 2\sqrt {1-{x^2}} + 1-{x^2}\,\,\,\,\,\,\, \Leftrightarrow \)
\( \Leftrightarrow \,\,\,\,\,\,\,2\sqrt {1-{x^2}} = 2{x^2}-1\,\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\,\,\left\{ {\begin{array}{*{20}{c}}{2{x^2}-1 \ge 0\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,}\\{4-4{x^2} = 4{x^4}-4{x^2} + 1}\end{array}} \right.\,\,\,\,\,\,\, \Leftrightarrow \)
\( \Leftrightarrow \,\,\,\,\,\,\,\left\{ {\begin{array}{*{20}{c}}{{x^2} \ge \dfrac{1}{2},}\\{4{x^4} = 3}\end{array}} \right.\,\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\,\,\left\{ {\begin{array}{*{20}{c}}{{x^2} \ge \dfrac{1}{2},\,\,\,\,\,\,\,\,\,}\\{\left[ {\begin{array}{*{20}{c}}{{x^2} = \dfrac{{\sqrt 3 }}{2},\,\,\,}\\{{x^2} = -\dfrac{{\sqrt 3 }}{2}\,}\end{array}} \right.}\end{array}} \right.\,\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\,\,{x^2} = \dfrac{{\sqrt 3 }}{2}\,\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\,\,x = \pm \dfrac{{\sqrt[4]{3}}}{{\sqrt 2 }}.\)
Ответ: \( \pm \dfrac{{\sqrt[4]{3}}}{{\sqrt 2 }}.\)