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9/11 : ( 6/7 - 5/6 )
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NT
1
17 tháng 3 2022
\(\frac{5}{8}\times\frac{9}{7}+\frac{5}{8}\times\frac{5}{7}\)
\(=\frac{5}{8}\times\left(\frac{9}{7}+\frac{5}{7}\right)\)
\(=\frac{5}{8}\times\frac{14}{7}\)
\(=\frac{5}{8}\times2\)
\(=\frac{10}{8}\)
\(=\frac{5}{4}\)
24 tháng 3 2021
\(A,\dfrac{7}{11}+\dfrac{5}{6}-\dfrac{-4}{11}-\dfrac{1}{6}\)
\(= (\dfrac{7}{11}+\dfrac{-4}{11})-(\dfrac{5}{6}-\dfrac{1}{6})\)
\(=\dfrac{3}{11}-\dfrac{2}{3}\)
\(= \dfrac{9}{33}-\dfrac{22}{33}\)
\(= \dfrac{-13}{33}\)
\(B,\dfrac{5}{9}.\dfrac{7}{3}+\dfrac{5}{9}.\dfrac{9}{13}-\dfrac{5}{9}:\dfrac{13}{5}\)
\(=(\dfrac{5}{9}.\dfrac{5}{9}:\dfrac{5}{9}).\dfrac{7}{3}+\dfrac{9}{13}-\dfrac{13}{5}\)
\(=(\dfrac{5}{9}.\dfrac{7}{3})+(\dfrac{9}{13}-\dfrac{13}{5})\)
\(=(\dfrac{5}{9}.\dfrac{21}{9})+(\dfrac{45}{65}-\dfrac{169}{65})\)
\(=\dfrac{35}{27}+\dfrac{-124}{65}\)
\(=\dfrac{2275}{1755}+\dfrac{-3348}{1755}\)
\(=\dfrac{-1073}{1755}\)
KV
28 tháng 6 2018
1 + 2 + 3 + 4 + 5 + 6 + 7 + 8 + 9 + 10 + 11 + 12
= ( 1 + 12 ) + ( 2 + 11 ) + ( 3 + 10 ) + ( 4 + 9 ) + ( 5 + 8 ) + ( 6 + 7 )
= 13 + 13 + 13 + 13 + 13 + 13
= 13 x 6
= 78
sai thui
18 tháng 8 2017
Bài 1:
\(A=\frac{5}{3.6}+\frac{5}{6.9}+....+\frac{5}{96.99}\)
\(\Rightarrow\frac{3}{5}A=\frac{3}{3.6}+\frac{3}{6.9}+....+\frac{3}{96.99}\)
\(\Rightarrow\frac{3}{5}A=\frac{1}{3}-\frac{1}{6}+\frac{1}{6}-\frac{1}{9}+...+\frac{1}{96}-\frac{1}{99}\)
\(=\frac{1}{3}-\frac{1}{99}=\frac{32}{99}\)
\(\Rightarrow A=\frac{32}{99}\div\frac{3}{5}=\frac{160}{297}\)
Bái 2:
\(B=\frac{2}{3.7}+\frac{2}{7.11}+...+\frac{2}{99.103}\)
\(\Rightarrow2B=\frac{4}{3.7}+\frac{4}{7.11}+....+\frac{4}{99.103}\)
\(\Rightarrow2B=\frac{1}{3}-\frac{1}{7}+\frac{1}{7}-\frac{1}{11}+....+\frac{1}{99}-\frac{1}{103}\)
\(=\frac{1}{3}-\frac{1}{103}=\frac{100}{309}\)
\(\Rightarrow B=\frac{100}{309}\div2=\frac{50}{309}\)
18 tháng 8 2017
Bài 1:
Ta có:
\(\frac{5}{n.\left(n+3\right)}=\frac{5}{3}.\frac{3}{n.\left(n+3\right)}=\frac{5}{3}.\frac{\left(n+3\right)-n}{n.\left(n+3\right)}=\frac{5}{3}.\left[\frac{n+3}{n.\left(n+3\right)}-\frac{n}{n\left(n+3\right)}\right]\)\(=\frac{5}{3}\left(\frac{1}{n}-\frac{1}{n+3}\right)\)
\(\frac{5}{3.6}+\frac{5}{6.9}+\frac{5}{9.12}+...+\frac{5}{96.99}=\frac{5}{3}\left(\frac{1}{3}-\frac{1}{6}+\frac{1}{6}-\frac{1}{9}+...+\frac{1}{96}-\frac{1}{99}\right)\)
Giải :
\(\frac{9}{11}:\left(\frac{6}{7}-\frac{5}{6}\right)\)
\(=\frac{9}{11}:\frac{1}{42}\)
\(=\frac{378}{11}\)