Монгол Бодлогын Сан

Заавар: $\dfrac{1}{2}=2^{-1}$ ба илтгэгч функцийн монотон чанар ашигла.

Бодолт: $\left(\dfrac{1}{2}\right)^{x^2-7} < 2^{3|x-1|}$ $(2^{-1})^{x^2-7} < 2^{3|x-1|}-x^2 +7 <3|x-1|$ $x^2+3|x-1| -7>0$ $$\left[\begin{array}{l} \left\{\begin{array}{c} x<1\\ x^2 -3(x -1) -7>0 \end{array}\right.\\ \left\{ \begin{array}{c} x\ge 1\\ (x^2+3(x-1) -7>0 \end{array}\right. \end{array}\right. \Rightarrow\left[\begin{array}{l} \left\{\begin{array}{c} x<1\\ x^2 -3x +3 -7>0 \end{array}\right.\\ \left\{ \begin{array}{c} x\ge 1\\ (x^2+3x-3 -7>0 \end{array}\right. \end{array}\right. \Rightarrow \left[\begin{array}{l} \left\{\begin{array}{c} x<1\\ (x^2-3x-4 >0 \end{array}\right.\\ \left\{ \begin{array}{c} x\ge 1\\ (x^2+3x-1) -10>0 \end{array}\right. \end{array}\right.$$ $$\left[\begin{array}{l} \left\{\begin{array}{c} x<1\\ (x -4) (x+1)>0 \end{array}\right.\\ \left\{ \begin{array}{c} x\ge 1\\ (x+5)(x-2) >0 \end{array}\right. \end{array}\right. \Rightarrow\left[ \begin{array}{c} x<-1\\ x>2 \end{array}\right. \Rightarrow$$ $\left]-\infty; {-1}\right[\cup]{2};\infty[$