a)
[tex]\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{3}{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{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{26}{3}+0 = -\frac{26}{3} = -8\frac{2}{3}[/tex]
b)
[tex]\frac{2}{3}-[0,8\cdot\frac{3}{(-2)^{2}}-\frac{(-2)^{2}}{5}]:(-1\frac{1}{2}) = \frac{2}{3}-(\frac{4}{5}\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{1}{5})\cdot(-\frac{2}{3}) = \frac{2}{3}-\frac{2}{15} = \frac{10}{15}-\frac{2}{15} = \frac{8}{15}[/tex]
