${\left(\frac{2}{3}\right)}^{-\frac{3}{2}}\cdot {\left(\frac{16}{81}\right)}^{\frac{1}{4}}-{\left(2,25\right)}^{\frac{1}{4}}:{\left(\frac{3}{2}\right)}^3=$  

$={\left(\frac{2}{3}\right)}^{-\frac{3}{2}}\cdot {\left({\left(\frac{2}{3}\right)}^4\right)}^{\frac{1}{4}}-{\left(2\frac{1}{4}\right)}^{\frac{1}{4}}:{\left({\left(\frac{2}{3}\right)}^{-1}\right)}^3=$     

$={\left(\frac{2}{3}\right)}^{-\frac{3}{2}}\cdot {\left(\frac{2}{3}\right)}^{4\cdot \frac{1}{4}}-{\left(\frac{9}{4}\right)}^{\frac{1}{4}}:{\left(\frac{2}{3}\right)}^{-3}=$  

$={\left(\frac{2}{3}\right)}^{-\frac{3}{2}}\cdot {\left(\frac{2}{3}\right)}^1-{\left({\left(\frac{2}{3}\right)}^{-2}\right)}^{\frac{1}{4}}:{\left(\frac{2}{3}\right)}^{-3}=$  

$={\left(\frac{2}{3}\right)}^{-\frac{3}{2}}\cdot {\left(\frac{2}{3}\right)}^1-{\left(\frac{2}{3}\right)}^{-\frac{1}{2}}:{\left(\frac{2}{3}\right)}^{-3}=$  

$={\left(\frac{2}{3}\right)}^{-\frac{3}{2}+1}-{\left(\frac{2}{3}\right)}^{-\frac{1}{2}-\left(-3\right)}={\left(\frac{2}{3}\right)}^{-\frac{1}{2}}-{\left(\frac{2}{3}\right)}^{\frac{5}{2}}=$  

$=\sqrt{\frac{3}{2}}-\sqrt{{\left(\frac{2}{3}\right)}^5}=\sqrt{\frac{3}{2}}-\sqrt{\frac{32}{243}}=$  

$=\frac{\sqrt{3}}{\sqrt{2}}-\frac{4\sqrt{2}}{9\sqrt{3}}=\frac{\sqrt{3}\cdot 9\sqrt{3}}{\sqrt{2}\cdot 9\sqrt{3}}-\frac{4\sqrt{2}\cdot \sqrt{2}}{9\sqrt{3}\cdot \sqrt{2}}=$  

$=\frac{27}{9\sqrt{6}}-\frac{8}{9\sqrt{6}}=\frac{19}{9\sqrt{6}}$  

Zatem pokazaliśmy, że jest to liczba dodatnia.