Odpowiedź:
[tex]Zad.10\\\\b)\\\\(2\frac{1}{6})&^2=(\frac{13}{6})^2=\frac{13}{6}\cdot\frac{13}{6}=\frac{169}{36}=4\frac{25}{36}\\\\\\(-1\frac{1}{9})^2=(-\frac{10}{9})^2=(-\frac{10}{9})\cdot(-\frac{10}{9})=\frac{100}{81}=1\frac{19}{81}\\\\\\(2\frac{1}{7})^2=(\frac{15}{7})^2=\frac{15}{7}\cdot\frac{15}{7}=\frac{225}{49}=4\frac{29}{49}[/tex]
[tex](-1\frac{1}{11})^2=(-\frac{12}{11})^2=(-\frac{12}{11})\cdot(-\frac{12}{11})=\frac{144}{121}=1\frac{23}{121}\\\\\\(3\frac{1}{3})^2=(\frac{10}{3})^2=\frac{10}{3}\cdot\frac{10}{3}=\frac{100}{9}=11\frac{1}{9}[/tex]
