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8) ((456.11 + 912).37) : 13 . 74
= [(456.11 + 456 . 2).37] : 13 . 74
= [456.(11 + 2).37] : 13 . 74
= 456 . 13 . 37 : 13 . 74
= 456 . 37 : 74
= 456 : ( 74 : 37 )
= 456 : 2
= 228
9) Ta có:
864 . 48 - 432 . 96
= 432 . 2 . 48 - 432 . 96
= 432 . 96 - 432 . 96
= 432 . ( 96 - 96 )
= 432 . 0
= 0
\(\Rightarrow\left(864.48-432.96\right):864.48.432\)
\(=0:864.48.432\)
\(=0\)
\(8,\left[\left(456.11+912\right).37\right]:13.74\)
\(=\left[\left(5016+912\right).37\right]:13.74\)
\(=\left(5928.37\right):13.74\)
\(=219336:13.74\)
\(=16872.74\)
\(=1248528\)
\(9,\left(864.48-432.96\right):864.48.432\)
\(=\left(41472-41472\right):864.48.432\)
\(=0:864.48.432\)
\(=0\)
Ta có: \(A=\frac{2020}{2021}+\frac{2021}{2022}\)
\(\Rightarrow A=\frac{2021}{2021}-\frac{1}{2021}+\frac{2022}{2022}-\frac{1}{2022}\)
\(\Rightarrow A=1-\frac{1}{2021}+1-\frac{1}{2022}\)
\(\Rightarrow A=1+1-\frac{1}{2021}-\frac{1}{2022}\)
\(\Rightarrow A=2-\frac{1}{2021}-\frac{1}{2022}\)
\(\Rightarrow A=2-\frac{1}{2021\cdot2022}\)
\(B=\frac{2020+2021}{2021+2022}\)
\(\Rightarrow B=\frac{2021+2022}{2021+2022}-\frac{2}{2021+2022}\)
\(\Rightarrow B=1-\frac{2}{2021+2022}\)
\(\Rightarrow B=1-\frac{2}{4043}\)
Vậy ta sẽ so sánh:
\(1-\frac{1}{2021\cdot2022};\frac{2}{4043}\)
Vì \(2021\cdot2022>4043\)nên \(\frac{1}{2021\cdot2022}< \frac{2}{4043}\)vậy \(1-\frac{1}{2021\cdot2022}>\frac{2}{4043}\)
\(\Rightarrow\frac{2020}{2021}+\frac{2021}{2022}>\frac{2020+2021}{2021+2022}\)
\(\Rightarrow A>B\)
Bác Hưng cần số xăng-ti-mét dây thép để làm móc treo là:
30 x 4 = 120 cm
Đáp số: 120 cm
\(a,8^4\times16^5\times32=\left(2^3\right)^4\times\left(2^4\right)^5\times2^5=2^{3\times4}\times2^{4\times5}\times2^5=2^{12}\times2^{20}\times2^5=2^{12+20+5}=2^{37}\)
\(b,27^4\times81^{10}=\left(3^3\right)^4\times\left(3^4\right)^{10}=3^{3\times4}\times3^{4\times10}=3^{12}\times3^{40}=3^{12+40}=3^{52}\)
\(c,625^5\div25^7=\left(5^4\right)^5\div\left(5^2\right)^7=5^{20}\div5^{14}=5^{20-14}=5^6\)
1) ( \(\frac{55}{3}\): 15 + \(\frac{26}{3}\) . \(\frac{7}{2}\)) : [(\(\frac{37}{3}\) + \(\frac{62}{7}\)) . \(\frac{7}{18}\)] : \(\frac{-1704}{445}\)
= ( \(\frac{55}{3}\). \(\frac{1}{15}\) + \(\frac{91}{3}\)) : [ \(\frac{445}{21}\) . \(\frac{7}{18}\)] . \(\frac{-445}{1704}\)
= ( \(\frac{11}{9}\)+ \(\frac{91}{3}\)) : \(\frac{445}{54}\). \(\frac{-445}{1704}\) = \(\frac{284}{9}\). \(\frac{54}{445}\). \(\frac{-445}{1704}\)= \(\frac{284}{9}\). (\(\frac{54}{445}\). \(\frac{-445}{1704}\))
= \(\frac{284}{8}\). \(\frac{-9}{284}\)
= \(\frac{-9}{8}\)
\(\frac{-1}{7}\left(9\frac{1}{2}-8,75\right)\div\frac{2}{7}+62,5\%\div1\frac{2}{3}\)
\(=\frac{-1}{7}\left(\frac{19}{2}-8,75\right).\frac{7}{2}+62,5\%\div\frac{5}{3}\)
\(=\frac{-1}{7}\left(\frac{19}{2}-\frac{875}{100}\right).\frac{7}{2}+62,5\%.\frac{3}{5}\)
\(=\frac{-1}{2}\left(\frac{38}{4}-\frac{35}{4}\right)+\frac{625}{100}.\frac{3}{5}\)
\(=\frac{-1}{2}.\frac{3}{4}+\frac{25}{4}.\frac{3}{5}\)
\(=\frac{-3}{8}+\frac{75}{20}\)
\(=\frac{-15}{40}+\frac{150}{40}\)
\(=\frac{135}{40}=\frac{27}{8}\)