c)
5,3+4\frac{5}{7}+\frac{1}{5}=5\frac{3}{10}+4\frac{5}{7}+\frac{1}{5}=5\frac{21}{70}+4\frac{50}{70}+\frac{14}{70}=9+\frac{85}{70}=9+1\frac{15}{70}=10\frac{15}{70}=10\frac{3}{14}
II sposób
5,3+4\frac{5}{7}+\frac{1}{5}=\frac{53}{10}+\frac{33}{7}+\frac{1}{5}=\frac{371}{70}+\frac{330}{70}+\frac{14}{70}=\frac{715}{70}=10\frac{15}{70}=10\frac{3}{14}
2)
10\frac{3}{4}-0,28+0,54=10,75-0,28+0,54=10.47+0,54=11,01
3)
9,72-3\frac{3}{5}*1\frac{2}{3}=9,72-\frac{\not18^6}{\not5^1}*\frac{\not5^1}{\not3^1}=9,72-6=3,72
4)
4,08-2,08:\frac{13}{35}=4,08-\frac{208}{100}*\frac{35}{13}=4\frac{8}{100}-\frac{\not52^4}{\not25^5}*\frac{\not35^7}{\not13^1}=-1,52
=4\frac{2}{25}-\frac{28}{5}=\frac{102}{25}-\frac{140}{25}=-\frac{38}{25}=-1\frac{13}{25}
5)
5,4:(5\frac{3}{7}-3,8)=5,4:(4\frac{10}{7}-3\frac{8}{10})=\frac{27}{5}:(4\frac{100}{70}-3\frac{56}{70})=\frac{27}{5}:1\frac{44}{70}=
=\frac{27}{5}:\frac{114}{70}=\frac{27}{5}:\frac{57}{35}=\frac{\not27^9}{\not5^1}*\frac{\not35^7}{\not57^{19}}=\frac{63}{19}=3\frac{6}{19}