Упр.823 ГДЗ Никольский Потапов 9 класс (Алгебра)

Задача

а) ((1,08-1,33)·18+0,6:2/15)/(20,1·0,1-2,1);
б) ((2,14-1,39)·1,2-0,75:5/6)/(12,1:0,1-1,2);
в) ((2 3/7-2 13/14)·(3 1/3)+5/51·17)/((3 5/7)·(2 1/2)-6 5/14);
г) ((4 1/5-5 1/2):(2 1/6)+303/304:505/304)/((7 1/5)·(5 1/2)-39,1).

Подробный ответ

а)

$$ \frac{(1{,}08-1{,}33)\cdot 18+0{,}6:\frac{2}{15}}{20{,}1\cdot 0{,}1-2{,}1} = \frac{-0{,}25\cdot 18+\frac{6}{10}\cdot \frac{15}{2}}{2{,}01-2{,}1} = \frac{-4{,}5+\frac{9}{2}}{-0{,}09} = \frac{-4{,}5+4{,}5}{-0{,}09} = \frac{0}{-0{,}09} =0. $$

б)

$$ \frac{(2{,}14-1{,}39)\cdot 1{,}2-0{,}75:\frac{5}{6}}{12{,}1:0{,}1-1{,}2} = \frac{0{,}75\cdot 1{,}2-\frac{75}{100}\cdot \frac{6}{5}}{121-1{,}2} = \frac{0{,}9-\frac{18}{20}}{119{,}8} = \frac{0{,}9-0{,}9}{119{,}8} = \frac{0}{119{,}8} =0. $$

в)

$$ \frac{\left(2\frac{3}{7}-2\frac{13}{14}\right)\cdot \left(3\frac{1}{3}\right)+\frac{5}{51}\cdot 17}{\left(3\frac{5}{7}\right)\cdot \left(2\frac{1}{2}\right)-6\frac{5}{14}} = \frac{\left(\frac{17}{7}-\frac{41}{14}\right)\cdot \frac{10}{3}+\frac{5}{3}}{\frac{26}{7}\cdot \frac{5}{2}-\frac{89}{14}} = \frac{\frac{34}{14}-\frac{41}{14}}{\frac{130}{14}-\frac{89}{14}}\cdot \frac{10}{3}+\frac{5}{3} $$
$$ = \frac{-\frac{7}{14}\cdot \frac{10}{3}+\frac{5}{3}}{\frac{41}{14}} = \frac{-\frac{1}{2}\cdot \frac{10}{3}+\frac{5}{3}}{\frac{41}{14}} = \frac{-\frac{5}{3}+\frac{5}{3}}{\frac{41}{14}} = \frac{0}{\frac{41}{14}} =0. $$

г)

$$ \frac{\left(4\frac{1}{5}-5\frac{1}{2}\right):2\frac{1}{6}+\frac{303}{304}:\frac{505}{304}}{\left(7\frac{1}{5}\right)\cdot \left(5\frac{1}{2}\right)-39{,}1} = \frac{\left(\frac{21}{5}-\frac{11}{2}\right):\frac{13}{6}+\frac{303}{304}\cdot \frac{304}{505}}{\frac{36}{5}\cdot \frac{11}{2}-39{,}1} $$
$$ = \frac{-\frac{13}{10}:\frac{13}{6}+\frac{3}{5}}{39{,}6-39{,}1} = \frac{-\frac{13}{10}\cdot \frac{6}{13}+\frac{3}{5}}{0{,}5} = \frac{-0{,}6+0{,}6}{0{,}5} = \frac{0}{0{,}5} =0. $$