[tex](-0,9)^2 : (\frac{3}{8})^2=(- \frac{9}{10})^2 :(\frac{3}{8})^2 =( \frac{\not{9}^3}{\not{10}^5}*\frac{\not{8}^4}{\not{3}^1})^2=\\\\=(-\frac{12}{5})^2=\frac{144}{25}=5,76\\\\\\\\(1\frac{2}{3})^4:(-2\frac{7}{9})^4=( \frac{5}{3})^4:(- \frac{25}{9})^4=( \frac{5}{3})^4*(- \frac{9}{25})^4=\\\\=( \frac{\not{5}^1}{\not{3}^1} * \frac{\not{9}^3}{\not{25}^5})^4=(\frac{3}{5})^4=\frac{81}{625}[/tex]
[tex]stosujemy\ wzor:\\\\a^n*b^n=(a*b)^n\\\\oraz\ to,\ ze:\\\\liczba\ ujemna\ podniesiona\ do\ potegi\ parzystej,\ jest\ liczba\ dodatnia[/tex]
