Jeśli a, b są dodatnimi liczbami rzeczywistymi, x, y dowolnymi liczbami rzeczywistymi, to  $a^x\cdot a^y=a^{x+y}$   $a^x:a^y=a^{x-y}$   ${\left(a^x\right)}^y=a^{x\cdot y}$   $a^x\cdot b^x={\left(a\cdot b\right)}^x$   $\frac{a^x}{b^x}={\left(\frac{a}{b}\right)}^x$

---

a)

Korzystając z praw działań na potęgach dostajemy 

$\sqrt{9}\cdot \left[{1,5}^{-1}+9^{-1,5}\right]-{27}^{-\frac{2}{3}}=$  

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

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

$=3\cdot \left[\frac{2}{3}+3^{-3}\right]-3^{-2}=$  

$=3\cdot \frac{2}{3}+3\cdot 3^{-3}-3^{-2}=$  

$=2+3^{1+\left(-3\right)}-3^{-2}=$  

$=2+\cancel{3^{-2}}-\cancel{3^{-2}}=2$   

---

b)

Korzystając z praw działań na potęgach dostajemy 

${\left[{\left(\frac{81}{625}\right)}^{-0,75}:{\left(1\frac{2}{3}\right)}^3-{0,125}^{\frac{1}{3}}\right]}^{-2}=$  

$={\left[{\left(\frac{81}{625}\right)}^{-\frac{3}{4}}:{\left(\frac{5}{3}\right)}^3-{\left(\frac{1}{8}\right)}^{\frac{1}{3}}\right]}^{-2}=$  

$={\left[{\left(\frac{3^4}{5^4}\right)}^{-\frac{3}{4}}:{\left(\frac{5}{3}\right)}^3-{\left(2^{-3}\right)}^{\frac{1}{3}}\right]}^{-2}=$  

$={\left[{\left({\left(\frac{3}{5}\right)}^4\right)}^{-\frac{3}{4}}:{\left({\left(\frac{3}{5}\right)}^{-1}\right)}^3-2^{-3\cdot \frac{1}{3}}\right]}^{-2}=$  

$={\left[{\left(\frac{3}{5}\right)}^{4\cdot \left(-\frac{3}{4}\right)}:{\left(\frac{3}{5}\right)}^{-3}-2^{-1}\right]}^{-2}=$  

$={\left[\underset{=1}{\underbrace{{\left(\frac{3}{5}\right)}^{-3}:{\left(\frac{3}{5}\right)}^{-3}}}-\frac{1}{2}\right]}^{-2}=$  

$={\left[1-\frac{1}{2}\right]}^{-2}={\left(\frac{1}{2}\right)}^{-2}=2^2=4$  

---

c)

Korzystając z praw działań na potęgach mamy 

$\sqrt[3]{0,375}\cdot \sqrt[3]{9}+{\left(3^{-1}-\sqrt[4]{\frac{16}{81}}\right)}^{-2}=$  

$=\sqrt[3]{\frac{3}{8}}\cdot \sqrt[3]{3^2}+{\left(\frac{1}{3}-\sqrt[4]{\frac{2^4}{3^4}}\right)}^{-2}=$  

$=\frac{\sqrt[3]{3}}{\sqrt[3]{8}}\cdot 3^{\frac{2}{3}}+{\left(\frac{1}{3}-\frac{2}{3}\right)}^{-2}=$  

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

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

$=\frac{3^1}{2}+9=\frac{3}{2}+9=10\frac{1}{2}$  

---

d)

Korzystając z praw działań na potęgach mamy 

${\left[\frac{{125}^{\frac{2}{3}}-{0,2}^{-1}}{{0,5}^{-2}}\cdot {0,2}^{-3}\right]}^{\frac{3}{4}}=$  

$={\left[\frac{{\left(5^3\right)}^{\frac{2}{3}}-{\left(\frac{1}{5}\right)}^{-1}}{{\left(\frac{1}{2}\right)}^{-2}}\cdot {\left(\frac{1}{5}\right)}^{-3}\right]}^{\frac{3}{4}}=$  

$={\left[\frac{5^{3\cdot \frac{2}{3}}-{\left(5^{-1}\right)}^{-1}}{2^2}\cdot 5^3\right]}^{\frac{3}{4}}=$  

$={\left[\frac{5^2-5}{4}\cdot 5^3\right]}^{\frac{3}{4}}=$  

$={\left[\frac{5\left(5-1\right)}{4}\cdot 5^3\right]}^{\frac{3}{4}}=$  

$={\left[\frac{5\cdot 4}{4}\cdot 5^3\right]}^{\frac{3}{4}}=$  

$={\left[5\cdot 5^3\right]}^{\frac{3}{4}}={\left(5^4\right)}^{\frac{3}{4}}=5^{4\cdot \frac{3}{4}}=5^3=125$