[tex]Mnozenie \ poteg\ o\ tym\ samym\ wykladniku \ polega\ na \ mnozeniu\ podstaw \\\\ a\ wykladnik\ pozostawiamy\ bez\ zmian .[/tex]
[tex]a^n*b^n=(a*b)^n \\\\d)\\\\2^3*3^3*(\frac{1}{6})^3=(2*3*\frac{1}{6})^3=(\not{6}^1*\frac{1}{\not{6}^1})^3=1^3=1 \\\\\\(\frac{5}{3})^4*(\frac{3}{4})^4*4^4=(\frac{5}{\not{3}^1}*\frac{\not{3}^1}{\not{4}^1}*\not{4}^1)^4=5^4=625\\\\\\2,5^5*4 ^5*0,2^5=(2,5*4 *0,2)^5 = (10*0,2)^5=2^5=32\\\\\\(1\frac{1}{7})^2*(\frac{1}{4})^2*(\frac{14}{3})^2=( \frac{8}{7})^2*(\frac{1}{4})^2*(\frac{14}{3})^2=( \frac{\not{8}^2}{\not{7}^1} *\frac{1} {\not{4}^1}*\frac{\not{14}^2}{3})^2=(\frac{4}{3})^2=\frac{16}{9}=1\frac{7}{9}[/tex]
