5.
[tex]b)\\\\\frac{\frac{2}{3}+\frac{1}{5}}{\frac{2}{5}-\frac{1}{2}}+(3\frac{2}{17}-4\frac{6}{13})\cdot(0,75-\frac{4}{4})=\\\\\\=\frac{\frac{10}{15}+\frac{3}{15}}{\frac{4}{10}-\frac{5}{10}}+(3\frac{2}{17}-4\frac{6}{13})\cdot(\frac{75}{100}-\frac{3}{4}=\\\\\\=\frac{\frac{13}{15}}{-\frac{1}{10}}+(3\frac{2}{17}-4\frac{6}{13})\cdot(\frac{3}{4}-\frac{3}{4})=\\\\\\=\frac{13}{15}\cdot(-\frac{10}{1})+(3\frac{2}{17}-4\frac{6}{13})\cdot0=\\\\\\=\frac{13}{3}\cdot(-2) + 0=\\\\\\=-\frac{26}{3} = -8\frac{2}{3}[/tex]
[tex]b)\\\\\frac{2}{3}-[0,8\cdot\frac{3}{(-2)^{2}}-\frac{(-2)^{2}}{5}]:(-1\frac{1}{2})=\frac{2}{3}-(\frac{8}{10}\cdot\frac{3}{4}-\frac{4}{5}):(-\frac{3}{2})=\frac{2}{3}-(\frac{3}{5}-\frac{4}{5})\cdot(-\frac{2}{3})=\\\\\\=\frac{2}{3}-(\frac{3}{5})\cdot(-\frac{2}{3}) = \frac{2}{3} - (-\frac{1}{5})\cdot(-\frac{2}{3})=\frac{2}{3}-\frac{2}{15} = \frac{10}{15}-\frac{2}{15} = \frac{8}{15}[/tex]
