Question

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Answer 1

$a)~\bigg(6,6\cdot\dfrac{40}{55} -\dfrac{19}{5}\bigg):0,5=\bigg(\dfrac{66}{10}\cdot\dfrac{40}{55}-\dfrac{19}{5}\bigg):\dfrac{5}{10} =\bigg(\dfrac{2640}{550}-\dfrac{19}{5} \bigg):\dfrac{5}{10}=\\\\=\dfrac{2640-2090}{550}\cdot\dfrac{10}{5} =\dfrac{550}{55}\cdot\dfrac{1}{5}=\dfrac{550}{275}=2$

$b)~\bigg(\dfrac{1}{3}+\dfrac{1}{3}\cdot2\bigg)\cdot\Big(14,9:149\Big)= \bigg(\dfrac{1}{3}+\dfrac{2}{3} \bigg)\cdot\bigg(\dfrac{149}{10}\cdot \dfrac{1}{149}\bigg)=\\\\= \dfrac{3}{3}\cdot\dfrac{1}{10}=1\cdot\dfrac{1}{10}=\dfrac{1}{10}$

$c)~\Big(1,05+0,3\cdot2,5\Big):1,1=\bigg(\dfrac{105}{100}+\dfrac{3}{10}\cdot\dfrac{25}{10}\bigg):\dfrac{11}{10}=\\\\=\bigg(\dfrac{105}{100} +\dfrac{75}{100}\bigg)\cdot\dfrac{10}{11}=\dfrac{180}{100}\cdot\dfrac{10}{11}=\dfrac{18}{11}$

$d)~\dfrac{1}{4}\cdot\bigg[\dfrac{17}{3}-4,(6)\bigg]\cdot2,5=\dfrac{1}{4}\cdot\bigg(\dfrac{17}{3}-\dfrac{46-4}{9} \bigg)\cdot\dfrac{25}{10}=\\\\=\dfrac{1}{4}\cdot\bigg(\dfrac{17}{3} -\dfrac{42}{9}\bigg)\cdot\dfrac{25}{10}=\dfrac{1}{4}\cdot\dfrac{17\cdot3-42}{9}\cdot\dfrac{25}{10}=\dfrac{1}{4}\cdot\dfrac{51-42}{9}\cdot\dfrac{25}{10} =\\\\=\dfrac{1}{4} \cdot\dfrac{9}{9}\cdot\dfrac{25}{10}=\dfrac{1}{4}\cdot1\cdot\dfrac{25}{10}=\dfrac{25}{40}=\dfrac{5}{8}$