Odpowiedź:
[tex]a)\ \ (\frac{7}{9})^2=\frac{7^2}{9^2}=\frac{49}{81}\\\\b)\ \ (\frac{1}{7})^2=\frac{1^2}{7^2}=\frac{1}{49}\\\\c)\ \ (\frac{3}{11})^2=\frac{3^2}{11^2}=\frac{9}{121}\\\\\\a)\ \ 3\frac{2}{3}:(1\frac{1}{2}+2\frac{3}{4})-\frac{5}{8}=\frac{11}{3}:(1\frac{2}{4}+2\frac{3}{4})-\frac{5}{8}=\frac{11}{3}:3\frac{5}{4}-\frac{5}{8}=\frac{11}{3}:\frac{17}{4}-\frac{5}{8}=\\\\=\frac{11}{3}\cdot\frac{4}{17}-\frac{5}{8}=\frac{44}{51}-\frac{5}{8}=\frac{352}{408}-\frac{255}{408}=\frac{97}{408}[/tex]
[tex]b)\ \ [2\frac{1}{3}+(5\frac{2}{5}-3\frac{2}{3})]\cdot1\frac{1}{4}=[2\frac{1}{3}+(5\frac{6}{15}-3\frac{10}{15})]\cdot\frac{5}{4}=[2\frac{1}{3}+(4\frac{21}{15}-3\frac{10}{15})]\cdot\frac{5}{4}=\\\\=(2\frac{1}{3}+1\frac{11}{15})\cdot\frac{5}{4}=(2\frac{5}{15}+1\frac{11}{15})\cdot\frac{5}{4}=3\frac{16}{15}\cdot\frac{5}{4}=\frac{61}{\not15_{3}}\cdot\frac{\not5^1}{4}=\frac{61}{12}=5\frac{1}{12}[/tex]
