Монгол Бодлогын Сан

Заавар: $\dfrac{1}{5}=5^{-1}$ ба илтгэгч функцийн монотон чанар ашигла.

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