Упр.8 Повторение ГДЗ Колмогоров 10-11 класс (Алгебра)
Найдите значения выражений:
а) $$\frac{2{,}75:1{,}1+3\frac{1}{3}}{2{,}5-0{,}4\cdot\left(-3\frac{1}{3}\right)}$$;
б) $$\frac{3\frac{1}{3}:10+0{,}175:\frac{7}{20}}{1\frac{3}{4}-1\frac{11}{17}\cdot\frac{51}{56}}$$;
в) $$\left(1{,}4-3{,}5:1\frac{1}{4}\right):2{,}4+3{,}4:2\frac{1}{8}$$;
г) $$\frac{1+\frac{1}{2}\cdot\frac{1}{0{,}25}}{6-\frac{46}{1+2{,}2\cdot10}}$$.
а) $$\frac{2{,}75:1{,}1+3\frac{1}{3}}{2{,}5-0{,}4\cdot\left(-3\frac{1}{3}\right)}=\frac{2{,}5+\frac{10}{3}}{2{,}5-\frac{2}{5}\cdot\left(-\frac{10}{3}\right)}=\frac{2{,}5+\frac{10}{3}}{2{,}5+\frac{4}{3}}$$
$$=\frac{\frac{15}{6}+\frac{20}{6}}{\frac{15}{6}+\frac{8}{6}}=\frac{\frac{35}{6}}{\frac{23}{6}}=\frac{35}{23}=1\frac{12}{23}.$$
б) $$\frac{3\frac{1}{3}:10+0{,}175:\frac{7}{20}}{1\frac{3}{4}-1\frac{11}{17}\cdot\frac{51}{56}}=\frac{\frac{10}{3}:10+0{,}175\cdot\frac{20}{7}}{\frac{7}{4}-\frac{28}{17}\cdot\frac{51}{56}}$$
$$=\frac{\frac{1}{3}+\frac{1}{2}}{\frac{7}{4}-\frac{3}{2}}=\frac{\frac{5}{6}}{\frac{1}{4}}=\frac{10}{3}=3\frac{1}{3}.$$
в) $$\left(1{,}4-3{,}5:1\frac{1}{4}\right):2{,}4+3{,}4:2\frac{1}{8}$$
$$=\left(1{,}4-3{,}5\cdot\frac{4}{5}\right):\frac{12}{5}+\frac{17}{5}:\frac{17}{8}$$
$$=(1{,}4-2{,}8)\cdot\frac{5}{12}+\frac{8}{5}=-1{,}4\cdot\frac{5}{12}+\frac{8}{5}$$
$$=-\frac{7}{5}\cdot\frac{5}{12}+\frac{8}{5}=-\frac{7}{12}+\frac{8}{5}=\frac{96-35}{60}=\frac{61}{60}=1\frac{1}{60}.$$
г) $$\frac{1+\frac{1}{2}\cdot\frac{1}{0{,}25}}{6-\frac{46}{1+2{,}2\cdot 10}}=\frac{1+\frac{1}{0{,}5}}{6-\frac{46}{1+22}}=\frac{1+2}{6-\frac{46}{23}}$$
$$=\frac{3}{6-2}=\frac{3}{4}.$$
Ответ
а) $$\frac{35}{23}$$; б) $$\frac{10}{3}$$; в) $$\frac{61}{60}$$; г) $$\frac{3}{4}$$.
