Предмет: Математика
Автор: annaklass04
Вопрос задан 2 года назад

Даю 95 баллов,за решение всех

Приложения:

Ответы

Ответ дал: flsh

1)
$\frac{12\frac{4}{5}*3\frac{3}{4}-4\frac{4}{11}*4\frac{1}{8}}{11\frac{2}{3}:2\frac{4}{7}}=\frac{\frac{64}{5}*\frac{15}{4}-\frac{48}{11}*\frac{33}{8}}{\frac{35}{3}:\frac{18}{7}}=\frac{\frac{16}{1}*\frac{3}{1}-\frac{6}{1}*\frac{3}{1}}{\frac{35}{3}*\frac{7}{18}}=\frac{48-18}{\frac{35}{3}*\frac{7}{18}}=\frac{30}{\frac{35}{3}*\frac{7}{18}}=$ $\frac{30}{1}*\frac{3}{35}*\frac{18}{7}=\frac{6}{1}*\frac{3}{7}*\frac{18}{7}=\frac{324}{49}=6\frac{30}{49}$
2)
$\frac{28\frac{4}{5}:13\frac{5}{7}+6\frac{3}{5}:\frac{2}{3}}{1\frac{11}{16}:2\frac{1}{4}}=\frac{\frac{144}{5}:\frac{96}{7}+\frac{33}{5}:\frac{2}{3}}{\frac{27}{16}:\frac{9}{4}}=\frac{\frac{144}{5}*\frac{7}{96}+\frac{33}{5}*\frac{3}{2}}{\frac{27}{16}*\frac{4}{9}}=\frac{\frac{3}{5}*\frac{7}{2}+\frac{33}{5}*\frac{3}{2}}{\frac{3}{4}}=$ $\frac{3*7+33*3}{5*2}*\frac{4}{3}=\frac{3*(7+33)}{5}*\frac{2}{3}=\frac{40}{5}*2=8*2=16$
3)
$\frac{2\frac{3}{8}:\frac{3}{4}+24\frac{7}{9}}{7\frac{1}{8}-157\frac{4}{5}:24}=\frac{\frac{19}{8}*\frac{4}{3}+\frac{223}{9}}{\frac{57}{8}-\frac{789}{5}*\frac{1}{24}}=\frac{\frac{19}{2}*\frac{1}{3}+\frac{223}{9}}{\frac{855-789}{8*3*5}}=\frac{\frac{57+446}{2*3*3}}{\frac{66}{8*3*5}}=\frac{\frac{503}{2*3*3}}{\frac{11}{4*5}}=\frac{503}{2*3*3}*\frac{4*5}{11}=$ $\frac{503}{3*3}*\frac{2*5}{11}=\frac{5030}{99}=50\frac{80}{99}$
4)
$\frac{(1 \frac{1}{2}+2 \frac{2}{3}+3 \frac{3}{4})*3 \frac{3}{5} }{14-15 \frac{1}{8}:2 \frac{1}{5} } =\frac{(\frac{3}{2}+\frac{8}{3}+\frac{15}{4})*\frac{18}{5} }{14-\frac{121}{8}:\frac{11}{5} }=\frac{\frac{18+32+45}{12}*\frac{18}{5} }{14-\frac{121}{8}*\frac{5}{11} }=\frac{\frac{95}{12}*\frac{18}{5} }{14-\frac{11}{8}*\frac{5}{1} }=\frac{\frac{19}{2}*\frac{3}{1} }{14-\frac{55}{8}}=$ $\frac{\frac{57}{2}}{\frac{112-55}{8}}=\frac{\frac{57}{2}}{\frac{57}{8}}=\frac{57}{2}*\frac{8}{57}=4$