Монгол Бодлогын Сан

Заавар: $$\sqrt{f(x)}>g(x)\Leftrightarrow\left\{\begin{array}{c}\left[\begin{array}{c}f(x)> g^2(x)\\ g(x)<0\end{array}\right.\\ f(x)\ge 0\end{array}\right.$$

Бодолт: $$\sqrt{x^2-x-6}>x-1\Leftrightarrow\left\{\begin{array}{c}\left[\begin{array}{c}x^2-x-6> (x-1)^2\\ x-1<0\end{array}\right.\\ x^2-x-6\ge 0\end{array}\right.$$ байна. $$\left[\begin{array}{c}x^2-x-6> (x-1)^2\\ x-1<0\end{array}\right.\Leftrightarrow \left[\begin{array}{c}x> 7\\ x<1\end{array}\right.$$ ба $x^2-x-6\ge 0\Leftrightarrow \left[\begin{array}{c}x\ge 3\\ x\le-2\end{array}\right.$ байна. Иймд тэнцэтгэл бишийн шийд нь $(-\infty;-2]\cup(7;+\infty)$ байна.