Question

Решать неравенства методом интервалов

Корень из (x^2 - 6x + 9) > 3

Answer 1

$\sqrt{x^2-6x+9} \ \textgreater \ 3\\\\ \sqrt{(x-3)^2}\ \textgreater \ 3\\\\|x-3|\ \textgreater \ 3 \; \; \Rightarrow \; \; \; \left [ {{x-3\ \textgreater \ 3} \atop {x-3\ \textless \ -3}} \right. \; \; \left [ {{x\ \textgreater \ 6} \atop {x\ \textless \ 0}} \right. \\\\//////\; (0)\; ---(6)\; //////\\\\x\in (-\infty ,0)\cup (6,+\infty )$