Все решения неравенства x во второй степени плюс корень из х во второй степни меньше 1/4,...

$x^2+\sqrt{x^2}\ \textless \ \frac{1}{4}, \\ x^2+|x|-\frac{1}{4}\ \textless \ 0, \\ \left [ {{x^2-x-\frac{1}{4}\ \textless \ 0,} \atop {x^2+x-\frac{1}{4}\ \textless \ 0,}} \right. \left [ {{4x^2-4x-1\ \textless \ 0,} \atop {4x^2+4x-1\ \textless \ 0,}} \right. \\ 4x^2\pm4x-1\ \textless \ 0, \\ D_1=(\pm2)^2-4\cdot(-1)=8, \\ x_{1,2}=\frac{2\pm\sqrt{8}}{4}=\frac{2\pm2\sqrt{2}}{4}=\frac{1\pm\sqrt{2}}{2}, \\ x_{3,4}=\frac{-2\pm\sqrt{8}}{4}=\frac{-2\pm2\sqrt{2}}{4}=\frac{-1\pm\sqrt{2}}{2},$
$\left [ {{\frac{1-\sqrt{2}}{2}\ \textless \ x\ \textless \ \frac{1+\sqrt{2}}{2},} \atop {\frac{-1-\sqrt{2}}{2}\ \textless \ x\ \textless \ \frac{-1+\sqrt{2}}{2};}} \right. \\ \frac{1-\sqrt{2}}{2}\ \textless \ x\ \textless \ \frac{-1+\sqrt{2}}{2}; \\ |\frac{-1+\sqrt{2}}{2}- \frac{1-\sqrt{2}}{2}|=\frac{-1+\sqrt{2}-1+\sqrt{2}}{2}=\frac{-2+2\sqrt{2}}{2}=\sqrt{2}-1\approx0,41$