a)

${\left(\frac{2}{3}\right)}^{-1}\le {\left(1,5\right)}^{|x+2|}$  

${\left(\frac{3}{2}\right)}^1\le {\left(\frac{3}{2}\right)}^{|x+2|}$      

$1\le \left|x+2\right|$  

$\left|x+2\right|\ge 1$  

$x+2\ge 1\vee x+2\le -1$  

$x\ge -1\vee x\le -3$  

$x\in (-\infty ,-3⟩\cup ⟨-1,+\infty )$  

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b)

${\left(\sqrt{2}\right)}^{|x-1|}<4$  

${\left(2^{\frac{1}{2}}\right)}^{|x-1|}<2^2$  

$2^{\frac{1}{2}|x-1|}<2^2$  

$\frac{1}{2}\left|x-1\right|<2\mid \cdot 2$  

$\left|x-1\right|<4$  

$x-1<4\wedge x-1>-4$  

$x<5\wedge x>-3$  

$x\in \left(-3,5\right)$  

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c)

${\left(\frac{1}{16}\right)}^{|x+2|}>{\left(2\sqrt[3]{2}\right)}^3$  

${\left(2^{-4}\right)}^{|x+2|}>{\left(2^1\cdot 2^{\frac{1}{3}}\right)}^3$  

$2^{-4|x+2|}>{\left(2^{\frac{4}{3}}\right)}^3$  

$2^{-4|x+2|}>2^4$  

$-4\left|x+2\right|>4\mid :\left(-1\right)$  

$\left|x+2\right|<-1$  

$x+2<-1\wedge x+2>1$  

$x<-3\wedge x>-1$  

$x\in \varnothing$  

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d)

${\left(0,4\right)}^{|x-6|}\le 2,5$  

${\left(\frac{4}{10}\right)}^{|x-6|}\le \frac{5}{2}$  

${\left(\frac{2}{5}\right)}^{|x-6|}\le {\left(\frac{2}{5}\right)}^{-1}$  

Podstawa potęgi  $\left(\frac{2}{5}\right)$   jest liczbą z przedziału (0;1), zatem znak nierówności zmieniamy na przeciwny:

$\left|x-6\right|\ge -1$  

$x-6\ge -1\vee x-6\le 1$  

$x\ge 5\vee x\le 7$  

$x\in \mathbb{R}$