[tex]36^{5}\cdot6^{7} = (6^{2})^{5}\cdot6^{7} = 6^{10}\cdot6^{7} =6\boxed{^{17}}\\\\\\9^{6}\cdot27^{5} = (3^{2})^{6}\cdot(3^{3})^{5} = 3^{12}\cdot3^{15} =3\boxed{^{27}}\\\\\\\frac{2^{14}\cdot5^{14}}{10^{12}} = \frac{(2\cdot5)^{14}}{10^{12}} = \frac{10^{14}}{10^{12}} = 10^{2} = \boxed{100}\\\\\\\frac{6^{17}\cdot5^{17}}{30^{17}} = \frac{(5\cdot6)^{17}}{30^{17}} = \frac{30^{17}}{30^{17}} = 30^{0} =\boxed{ 1}[/tex]
Wykorzystano własności potęg:
[tex](a^{m})^{n} = a^{m\cdot n}\\\\a^{m}\cdot a^{n} = a^{m+n}\\\\a^{m}:a^{n} = a^{m-n}\\\\a^{n}\cdot b^{n} = (a\cdot b)^{n}[/tex]
[tex]a^{0} = 1[/tex]
