\({x^2}-2a\left( {x-1} \right)-1 = 0\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\,\left( {x-1} \right)\left( {x + 1} \right)-2a\left( {x-1} \right) = 0\,\,\,\,\,\, \Leftrightarrow \)
\( \Leftrightarrow \,\,\,\,\,\,\,\left( {x-1} \right)\left( {x + 1-2a} \right) = 0\,\,\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\,\,\,\left[ {\begin{array}{*{20}{c}}{x = 1,\,\,\,\,\,\,\,\,\,\,\,\,}\\{x = 2a-1.}\end{array}} \right.\)
По условию сумма корней уравнения должна быть равна сумме квадратов этих корней:
\({x_1} + {x_2} = x_1^2 + x_2^2\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\,1 + 2a-1 = 1 + {\left( {2a-1} \right)^2}\,\,\,\,\,\, \Leftrightarrow \)
\( \Leftrightarrow \,\,\,\,\,\,\,2{a^2}-3a + 1 = 0\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\,\left[ {\begin{array}{*{20}{c}}{a = 1,\,\,}\\{a = \dfrac{1}{2}.}\end{array}} \right.\)
ОТВЕТ: 1/2; 1.