Odpowiedź:
[tex]c)\ \ (4\frac{1}{2})^3\cdot(-\frac{2}{3})^3\cdot(-2)^3=(\frac{9}{2})^3\cdot((-\frac{2}{3})\cdot(-2))^3=(\frac{9}{2}\cdot\frac{2}{3}\cdot2)^3=\(\frac{\not9^3}{\not2_{1}}\cdot\frac{\not2^1}{\not3_{1}}\cdot2)^3=\\\\=(3\cdot2)^3=6^3=216\\\\\\d)\ \ (2\frac{1}{2})^4\cdot0,6^4:(-15)^4=(\frac{5}{2})^4\cdot(\frac{3}{5})^4:15^4=(\frac{\not5^1}{2}\cdot\frac{3}{\not5_{1}}:15)^4=(\frac{\not3^1}{2}\cdot\frac{1}{\not15_{5}})^4=\\\\=(\frac{1}{2}\cdot\frac{1}{5})^4=(\frac{1}{10})^4=\frac{1}{10000}[/tex]
[tex](-\frac{2}{3})\cdot(-2)\ \ \ \ iloczyn\ \ dw\'och\ \ liczb\ \ ujemnych\ \ jest\ \ dodatni\\\\(-15)^4\ \ \ \ \ \ ujemna\ \ podstawa\ \ podniesiona\ \ do\ \ parzystej\ \ potegi\ \ daje\ \ wynik\ \ dodatni[/tex]
