[tex]\displaystyle\bf\\a)\\\\\frac{\left(4\dfrac{3}{5}-2\dfrac{1}{5}\right)\times\dfrac{5}{6}+3}{\dfrac{18}{15}:\dfrac{6}{5}}=\\\\\\=\frac{\left(\dfrac{4\times5+3}{5}-\dfrac{2\times5+1}{5}\right)\times\dfrac{5}{6}+3}{\dfrac{18}{15}\times\dfrac{5}{6}}=\\\\\\=\frac{\left(\dfrac{23}{5}-\dfrac{11}{5}\right)\times\dfrac{5}{6}+3}{\dfrac{18\times5}{15\times6}}=\\\\\\=\frac{\dfrac{23-11}{5}\times\dfrac{5}{6}+3}{\dfrac{90}{90}}=\\\\\\=\frac{\dfrac{12}{5}\times\dfrac{5}{6}+3}{1}=\\\\=2+3=\boxed{\bf5}[/tex]
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[tex]\displaystyle\bf\\b)\\\\ \left(\frac{4}{5}\right)^2\times\frac{20}{8}-\frac{9}{5}:3\frac{6}{10}+\frac{1}{10}=\\\\\\=\frac{4^2}{5^2}\times\frac{20}{8}-\frac{9}{5}:\frac{3\times10+6}{10}+\frac{1}{10}= \\\\\\=\frac{16}{25}\times\frac{20}{8}-\frac{9}{5}:\frac{36}{10}+\frac{1}{10}=[/tex]
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[tex]\displaystyle\bf\\=\frac{16\times20}{25\times8}-\frac{9}{5}\times\frac{10}{36}+\frac{1}{10}=\\\\\\=\frac{2\times4}{5}-\frac{2}{4}+\frac{1}{10}=\\\\\\=\frac{8}{5}-\frac{2}{4}+\frac{1}{10}=\\\\\\=\frac{32}{20}-\frac{10}{20}+\frac{2}{20}=\\\\\\=\frac{32-10+2}{20}=\frac{22+2}{20}=\frac{24}{20}=\frac{6}{5}=\boxed{\bf1\frac{1}{5}}[/tex]
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[tex]\displaystyle\bf\\\\c)\\\\\frac{5}{8}-a~parte~din~248=\\\\\\=\frac{5}{8}\times248=\frac{5\times31}{1}=5\times31=\boxed{\bf155}\\\\\bullet\\\\d)\\\\34\%~din~150=\\\\=34\%\times150=\\\\=\frac{34}{100}\times150=\\\\\\=\frac{17}{50}\times150=\\\\\\=\frac{17\times150}{50}=17\times3=\boxed{\bf51}[/tex]
