\(\sqrt x + \sqrt {x + \sqrt {1-x} } = 1\,\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\,\,\sqrt {x + \sqrt {1-x} } = 1-\sqrt x \,\,\,\,\,\,\, \Leftrightarrow \)
\( \Leftrightarrow \,\,\,\,\,\,\,\left\{ {\begin{array}{*{20}{c}}{1-\sqrt x \ge 0,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,}\\{x + \sqrt {1-x} = 1-2\sqrt x + x}\end{array}} \right.\,\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\,\,\left\{ {\begin{array}{*{20}{c}}{\sqrt x \le 1,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,}\\{\sqrt {1-x} = 1-2\sqrt x }\end{array}} \right.\,\,\,\,\,\,\, \Leftrightarrow \)
\( \Leftrightarrow \,\,\,\,\,\,\,\left\{ {\begin{array}{*{20}{c}}{\sqrt x \le 1,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,}\\{1-2\sqrt x \ge 0,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,}\\{1-x = 1-4\sqrt x + 4x}\end{array}} \right.\,\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\,\,\left\{ {\begin{array}{*{20}{c}}{\sqrt x \le 1,\,\,\,\,\,\,}\\{\sqrt x \le \dfrac{1}{2},\,\,\,\,}\\{4\sqrt x = 5x}\end{array}} \right.\,\,\,\,\,\,\, \Leftrightarrow \)
\( \Leftrightarrow \,\,\,\,\,\,\,\left\{ {\begin{array}{*{20}{c}}{0 \le x \le \frac{1}{4},\,\,}\\{16x = 25{x^2}}\end{array}} \right.\,\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\,\,\left\{ {\begin{array}{*{20}{c}}{0 \le x \le \dfrac{1}{4},}\\{\left[ {\begin{array}{*{20}{c}}{x = 0,\,\,}\\{x = \dfrac{{16}}{{25}}}\end{array}} \right.}\end{array}} \right.\,\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\,\,x = 0.\)
Ответ: 0.