23 và1/4.7/5-13 và1/4:5/7 lm giup nhen
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LP
1
27 tháng 8 2021
Ta có: \(C=5\dfrac{9}{10}:\dfrac{3}{2}-\left(2\dfrac{1}{3}\cdot4\dfrac{1}{2}-2\cdot2\dfrac{1}{3}\right):\dfrac{7}{4}\)
\(=\dfrac{59}{10}\cdot\dfrac{2}{3}-\left(\dfrac{7}{3}\cdot\dfrac{9}{2}-2\cdot\dfrac{7}{3}\right):\dfrac{7}{4}\)
\(=\dfrac{59}{15}-\left(\dfrac{7}{3}\cdot\dfrac{5}{2}\right):\dfrac{7}{4}\)
\(=\dfrac{59}{15}-\dfrac{35}{6}\cdot\dfrac{4}{7}\)
\(=\dfrac{59}{15}-\dfrac{10}{3}\)
\(=\dfrac{3}{5}\)
28 tháng 6 2021
`a)1<3`
`=>1/5<3/5`
`b)21>9`
`=>8/21<8/9`
`c)3/5<5/5=1`
`d)7/5>5/5=1`
PT
1
27 tháng 7 2023
\(\dfrac{3\dfrac{1}{3}+\dfrac{1}{5}+2\dfrac{7}{15}}{\left(\dfrac{3}{10}+\dfrac{1}{4}+\dfrac{7}{20}\right).\dfrac{5}{6}}=\dfrac{\dfrac{10}{3}+\dfrac{1}{5}+\dfrac{37}{15}}{\left(\dfrac{6}{20}+\dfrac{5}{20}+\dfrac{7}{20}\right).\dfrac{5}{6}}\)
\(=\dfrac{\dfrac{50}{15}+\dfrac{3}{15}+\dfrac{37}{15}}{\dfrac{18}{20}.\dfrac{5}{6}}=\dfrac{6}{\dfrac{3}{4}}=\dfrac{6.4}{3}=8\)
HH
8 tháng 8 2020
Bài 1 :
\(a)\frac{-17}{30}-\frac{11}{-15}+\left(-\frac{7}{12}\right)\)
\(=\frac{1}{6}+\left(-\frac{7}{12}\right)\)
\(=-\frac{5}{12}\)
\(b)-\frac{5}{9}+\frac{5}{9}:\left(1\frac{2}{3}-2\frac{1}{12}\right)\)
\(=-\frac{5}{9}+\frac{5}{9}:\left(\frac{5}{3}-\frac{25}{12}\right)\)
\(=-\frac{5}{9}+\frac{5}{9}:\left(-\frac{5}{12}\right)\)
\(=-\frac{5}{9}+\left(-\frac{2}{3}\right)\)
\(=-\frac{1}{9}\)
\(c)-\frac{7}{25}\times\frac{11}{13}+\left(-\frac{7}{25}\right)\times\frac{2}{13}-\frac{18}{25}\)
\(=-\frac{77}{325}+\left(-\frac{14}{325}\right)-\frac{18}{25}\)
\(=-\frac{7}{25}-\frac{18}{25}\)
\(=-1\)
DV
0
HG
4 tháng 2 2021
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NL
0
LT
1
DD
Đoàn Đức Hà
Giáo viên
6 tháng 6 2021
\(\frac{4}{3}+\frac{9}{8}+...+\frac{9801}{9800}\)
\(=1+\frac{1}{2^2-1}+1+\frac{1}{3^2-1}+...+1+\frac{1}{99^2-1}\)
\(=98+\frac{1}{1.3}+\frac{1}{2.4}+\frac{1}{3.5}+\frac{1}{4.6}+...+\frac{1}{97.99}+\frac{1}{98.100}\)
\(=98+\frac{1}{2}\left(\frac{2}{1.3}+\frac{2}{2.4}+\frac{2}{3.5}+\frac{2}{4.6}+...+\frac{2}{97.99}+\frac{2}{98.100}\right)\)
\(=98+\frac{1}{2}\left(\frac{3-1}{1.3}+\frac{4-2}{2.4}+\frac{5-3}{3.5}+\frac{6-4}{4.6}+...+\frac{99-97}{97.99}+\frac{100-98}{98.100}\right)\)
\(=98+\frac{1}{2}\left(1-\frac{1}{3}+\frac{1}{2}-\frac{1}{4}+\frac{1}{3}-\frac{1}{5}+\frac{1}{4}-\frac{1}{6}+...+\frac{1}{97}-\frac{1}{99}+\frac{1}{98}-\frac{1}{100}\right)\)
\(=98+\frac{1}{2}\left(1+\frac{1}{2}-\frac{1}{99}-\frac{1}{100}\right)\)
\(=98+\frac{14651}{19800}\)
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