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\(a,8^4\times16^5\times32=\left(2^3\right)^4\times\left(2^4\right)^5\times2^5=2^{3\times4}\times2^{4\times5}\times2^5=2^{12}\times2^{20}\times2^5=2^{12+20+5}=2^{37}\)
\(b,27^4\times81^{10}=\left(3^3\right)^4\times\left(3^4\right)^{10}=3^{3\times4}\times3^{4\times10}=3^{12}\times3^{40}=3^{12+40}=3^{52}\)
\(c,625^5\div25^7=\left(5^4\right)^5\div\left(5^2\right)^7=5^{20}\div5^{14}=5^{20-14}=5^6\)
1) ( \(\frac{55}{3}\): 15 + \(\frac{26}{3}\) . \(\frac{7}{2}\)) : [(\(\frac{37}{3}\) + \(\frac{62}{7}\)) . \(\frac{7}{18}\)] : \(\frac{-1704}{445}\)
= ( \(\frac{55}{3}\). \(\frac{1}{15}\) + \(\frac{91}{3}\)) : [ \(\frac{445}{21}\) . \(\frac{7}{18}\)] . \(\frac{-445}{1704}\)
= ( \(\frac{11}{9}\)+ \(\frac{91}{3}\)) : \(\frac{445}{54}\). \(\frac{-445}{1704}\) = \(\frac{284}{9}\). \(\frac{54}{445}\). \(\frac{-445}{1704}\)= \(\frac{284}{9}\). (\(\frac{54}{445}\). \(\frac{-445}{1704}\))
= \(\frac{284}{8}\). \(\frac{-9}{284}\)
= \(\frac{-9}{8}\)
\(\frac{-7}{12}:\frac{13}{6}+\frac{-7}{12}:\frac{13}{7}.\frac{2.|-8|}{3}\)
\(=\frac{-7}{12}.\frac{6}{13}+\frac{-7}{12}.\frac{7}{13}.\frac{2.8}{3}\)
\(=\frac{-7}{12}.\left(\frac{6}{13}+\frac{7}{13}.\frac{2.8}{3}\right)\)
\(=\frac{-7}{12}.\frac{10}{3}\)
\(=\frac{-35}{18}\)
\(\frac{-7}{12}:\frac{13}{6}+\frac{-7}{12}:\frac{13}{7}\times\frac{2\times\left|-8\right|}{3}\)
\(=\frac{-7}{12}\times\frac{6}{13}+\frac{-7}{12}\times\frac{7}{13}\times\frac{2\times8}{3}\)
\(=\frac{-7}{12}\times\left(\frac{6}{13}+\frac{7}{13}+\frac{2\times8}{3}\right)\)
\(=\frac{-7}{12}\times\frac{10}{3}\)
\(=\frac{-35}{18}\)
Rất vui khi giúp đc bạn.<3. Nếu có sai sót mong bạn bỏ qua
\(B=\left(1-\frac{1}{2}\right)\cdot\left(1-\frac{1}{3}\right)\cdot\left(1-\frac{1}{4}\right)\cdot...\cdot\left(1-\frac{1}{20}\right)\)
\(B=\frac{1}{2}\cdot\frac{2}{3}\cdot\frac{3}{4}\cdot...\cdot\frac{19}{20}\)
\(B=\frac{1}{20}\)
\(\frac{5}{9}\cdot\frac{7}{13}+\frac{5}{9}\cdot\frac{9}{13}+\frac{3}{13}\cdot\left(-\frac{5}{9}\right)\)
\(=\frac{5}{9}\cdot\frac{7}{13}+\frac{5}{9}\cdot\frac{9}{13}-\frac{3}{13}\cdot\frac{5}{9}\)
\(=\frac{5}{9}\cdot\left(\frac{7}{13}+\frac{9}{13}-\frac{3}{13}\right)\)
\(=\frac{5}{9}\)