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a) \(A=49\frac{8}{23}-\left(5\frac{7}{32}+14\frac{8}{23}\right)\)
\(A=49\frac{8}{23}-5\frac{7}{32}-14\frac{8}{23}\)
\(A=\left(49\frac{8}{23}-14\frac{8}{23}\right)-5\frac{7}{32}=35-5\frac{7}{32}=35-\frac{167}{32}=\frac{953}{32}\)
b) \(B=\frac{-3}{7}\cdot\frac{5}{9}+\frac{4}{9}:\frac{-7}{3}+2\frac{3}{7}\)
\(B=\frac{-3}{7}\cdot\frac{5}{9}+\frac{4}{9}\cdot\frac{-3}{7}+2\frac{3}{7}\)
\(B=\frac{-3}{7}\left(\frac{5}{9}+\frac{4}{9}\right)+2\frac{3}{7}\)
\(B=\frac{-3}{7}+\frac{17}{7}=\frac{14}{7}=2\)
c) \(C=\left(19\frac{5}{8}:\frac{7}{12}-13\frac{1}{4}:\frac{7}{12}\right)\cdot\frac{4}{5}\)
\(C=\left[\left(19\frac{5}{8}-13\frac{1}{4}\right):\frac{7}{12}\right]\cdot\frac{4}{5}\)
\(C=\left[\left(19\frac{5}{8}-13\frac{2}{8}\right):\frac{7}{12}\right]\cdot\frac{4}{5}\)
\(C=6\frac{3}{8}\cdot\frac{4}{5}=\frac{51}{8}\cdot\frac{4}{5}=\frac{51}{2}\cdot\frac{1}{5}=\frac{51}{10}\)
d) \(D=\frac{54\cdot107-53}{53\cdot107+54}=\frac{\left(53+1\right)\cdot107-53}{53\cdot107+54}=\frac{53\cdot107+107-53}{53\cdot107+54}=\frac{53\cdot107+54}{53\cdot107+54}=1\)
1) ( \(\frac{55}{3}\): 15 + \(\frac{26}{3}\) . \(\frac{7}{2}\)) : [(\(\frac{37}{3}\) + \(\frac{62}{7}\)) . \(\frac{7}{18}\)] : \(\frac{-1704}{445}\)
= ( \(\frac{55}{3}\). \(\frac{1}{15}\) + \(\frac{91}{3}\)) : [ \(\frac{445}{21}\) . \(\frac{7}{18}\)] . \(\frac{-445}{1704}\)
= ( \(\frac{11}{9}\)+ \(\frac{91}{3}\)) : \(\frac{445}{54}\). \(\frac{-445}{1704}\) = \(\frac{284}{9}\). \(\frac{54}{445}\). \(\frac{-445}{1704}\)= \(\frac{284}{9}\). (\(\frac{54}{445}\). \(\frac{-445}{1704}\))
= \(\frac{284}{8}\). \(\frac{-9}{284}\)
= \(\frac{-9}{8}\)
\(\frac{7}{13}.\frac{7}{15}-\frac{5}{12}.\frac{21}{39}+\frac{49}{91}.\frac{8}{15}\\ =\frac{7}{13}.\frac{7}{15}-\frac{5}{12}.\frac{7}{13}+\frac{7}{13}.\frac{8}{15}\\ =\frac{7}{13}\left(\frac{7}{15}+\frac{8}{15}-\frac{5}{12}\right)\\ =\frac{7}{13}\left(1-\frac{5}{12}\right)\\ =\frac{7}{13}.\frac{7}{12}\\ =\frac{49}{156}\)
\(\frac{7}{13}.\frac{7}{15}-\frac{5}{12}.\frac{21}{39}+\frac{49}{91}.\frac{8}{15}\\ =\frac{7}{13}.\frac{7}{15}-\frac{5}{12}.\frac{7}{13}+\frac{7}{13}.\frac{8}{15}\\ =\frac{7}{13}\left(\frac{7}{15}-\frac{5}{12}+\frac{8}{15}\right)\\ =\frac{7}{13}\left(\frac{7}{15}+\frac{8}{15}-\frac{5}{12}\right)\\ =\frac{7}{13}\left(1-\frac{5}{12}\right)\\ =\frac{7}{13}.\frac{712}{ }\)
\(\frac{7}{13}.\frac{7}{12}=\frac{49}{156}\)
\(\frac{5}{9}\cdot\frac{7}{13}+\frac{5}{9}\cdot\frac{9}{13}+\frac{3}{13}\cdot\left(-\frac{5}{9}\right)\)
\(=\frac{5}{9}\cdot\frac{7}{13}+\frac{5}{9}\cdot\frac{9}{13}-\frac{3}{13}\cdot\frac{5}{9}\)
\(=\frac{5}{9}\cdot\left(\frac{7}{13}+\frac{9}{13}-\frac{3}{13}\right)\)
\(=\frac{5}{9}\)
\(\frac{-7}{12}:\frac{13}{6}+\frac{-7}{12}:\frac{13}{7}.\frac{2.|-8|}{3}\)
\(=\frac{-7}{12}.\frac{6}{13}+\frac{-7}{12}.\frac{7}{13}.\frac{2.8}{3}\)
\(=\frac{-7}{12}.\left(\frac{6}{13}+\frac{7}{13}.\frac{2.8}{3}\right)\)
\(=\frac{-7}{12}.\frac{10}{3}\)
\(=\frac{-35}{18}\)
\(\frac{-7}{12}:\frac{13}{6}+\frac{-7}{12}:\frac{13}{7}\times\frac{2\times\left|-8\right|}{3}\)
\(=\frac{-7}{12}\times\frac{6}{13}+\frac{-7}{12}\times\frac{7}{13}\times\frac{2\times8}{3}\)
\(=\frac{-7}{12}\times\left(\frac{6}{13}+\frac{7}{13}+\frac{2\times8}{3}\right)\)
\(=\frac{-7}{12}\times\frac{10}{3}\)
\(=\frac{-35}{18}\)
Rất vui khi giúp đc bạn.<3. Nếu có sai sót mong bạn bỏ qua