Odpowiedź:
[tex]c)\ \ 2\frac{1}{3}\cdot\frac{3}{7}+\frac{2}{3}:3\frac{1}{3}+\frac{1}{15}=\frac{\not7^1}{\not3_{1} }\cdot\frac{\not3^1}{\not7_{1}}+\frac{2}{3}:\frac{10}{3}+\frac{1}{15}=1+\frac{\not2^1}{\not3_{1}}\cdot\frac{\not3^1}{\not10_{5}}+\frac{1}{15}=1+\frac{1}{5}+\frac{1}{15}=\\\\=\frac{15}{15}+\frac{3}{15}+\frac{1}{15}=\frac{19}{15}=1\frac{4}{15}[/tex]
[tex]c)\ \ 2\frac{2}{3}\cdot1\frac{1}{5}:(7\frac{8}{21}-\frac{3}{7}\cdot11)=\frac{8}{\not3_{1} }\cdot\frac{\not6^2}{5}:(7\frac{8}{21}-\frac{33}{7})=\frac{16}{5}:(\frac{155}{21}-\frac{99}{21})=\frac{16}{5}:\frac{56}{21}=\\\\=\frac{\not16^2}{5}\cdot\frac{21}{\not56_{7}}=\frac{42}{35}=\frac{6}{5}=1\frac{1}{5}[/tex]
[tex]c)\ \ 1\frac{3}{4}\cdot(3\frac{2}{3}+4\frac{1}{5}+1\frac{1}{3}):1\frac{9}{14}\cdot\frac{1}{7}=\frac{7}{4}\cdot(5+4\frac{1}{5}):\frac{23}{14}\cdot\frac{1}{7}=\frac{7}{4}\cdot9\frac{1}{5}:\frac{23}{14}\cdot\frac{1}{7}=\\\\=\frac{7}{4}\cdot\frac{\not46^2}{5}\cdot\frac{\not14^2}{\not23_{1}}\cdot\frac{1}{\not7_{1}}=\frac{28}{20}=\frac{7}{5}=1\frac{2}{5}[/tex]
