Монгол Бодлогын Сан

Заавар: Логарифмийн суурь нь 1-ээс бага тул $$\log_{\frac12}(x^2+x-2)>-2\Leftrightarrow \left\{\begin{array}{c}x^2+x-2<\left(\dfrac12\right)^{-2}=4\\x^2+x-2>0\end{array}\right.$$ байна.

Бодолт: $$\log_{\frac12}(x^2+x-2)>-2\Leftrightarrow \left\{\begin{array}{c}x^2+x-2<\left(\dfrac12\right)^{-2}=4\\x^2+x-2>0\end{array}\right.\Leftrightarrow$$ $$\left\{\begin{array}{c}x^2+x-6<0\\x^2+x-2>0\end{array}\right.\Leftrightarrow \left\{\begin{array}{c}(x+3)(x-2)<0\\(x+2)(x-1)>0\end{array}\right.\Leftrightarrow\left\{\begin{array}{c}-3< x<2\\x<-2\cup x>1\end{array}\right.$$ тул $x\in (-3;-2)\cup(1;2)$ байна.