Odpowiedź:
[tex]b)\ \ \dfrac{(2-\frac{1}{7})\cdot3}{2+0,6}-8\cdot\frac{3}{7}=\dfrac{(\frac{14}{7}-\frac{1}{7})\cdot3}{2,6}-\frac{24}{7}=\dfrac{\frac{13}{7}\cdot3}{\frac{26}{10}}-\frac{24}{7}=\dfrac{\frac{39}{7}}{\frac{13}{5}}-\frac{24}{7}=\frac{39}{7}:\frac{13}{5}-\frac{24}{7}=\\\\\\=\frac{\not39^3}{7}\cdot\frac{5}{\not13_{1}}-\frac{24}{7}=\frac{15}{7}-\frac{24}{7}=-\frac{9}{7}=-1\frac{2}{7}[/tex]
[tex]e)\ \ [(-\frac{2}{3})^2:\frac{8}{9}]^2-(\frac{3}{4}-1\frac{1}{8})=(\frac{\not4^1}{\not3_{1} }\cdot\frac{\not9^3}{\not8_{2}})^2-(\frac{3}{4}-\frac{9}{8})=(\frac{1}{2})^2-(\frac{6}{8}-\frac{9}{8})=\frac{1}{4}-(-\frac{3}{8})=\\\\=\frac{1}{4}+\frac{3}{8}=\frac{2}{8}+\frac{3}{8}=\frac{5}{8}[/tex]
[tex]f)\ \ 1,6\cdot\frac{7}{16}-\frac{2,7}{3^3}-\frac{5\cdot0,7}{0,35}=\frac{\not16}{10}\cdot\frac{7}{\not16}-\frac{2,7}{27}-\frac{3,5}{0,35}=\frac{7}{10}-0,1-10=0,7-0,1-10=\\\\=0,6-10=-9,4[/tex]
