Bài 9: Tính...
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Bài 9: Tính nhanh
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30 tháng 8 2021
a: Ta có: \(A=\dfrac{1}{1\cdot3}+\dfrac{1}{3\cdot5}+...+\dfrac{1}{2009\cdot2011}\)
\(=\dfrac{1}{2}\left(\dfrac{2}{1\cdot3}+\dfrac{2}{3\cdot5}+...+\dfrac{2}{2009\cdot2011}\right)\)
\(=\dfrac{1}{2}\left(1-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+...+\dfrac{1}{2009}-\dfrac{1}{2011}\right)\)
\(=\dfrac{1}{2}\cdot\dfrac{2010}{2011}=\dfrac{1005}{2011}\)
b: Ta có: \(B=\left(1+\dfrac{1}{2}\right)\left(1+\dfrac{1}{3}\right)+...+\left(1+\dfrac{1}{100}\right)\)
\(=\dfrac{3}{2}\cdot\dfrac{4}{3}\cdot...\cdot\dfrac{101}{100}\)
\(=\dfrac{101}{2}\)
HP
30 tháng 8 2021
\(A=\dfrac{1}{1.3}+\dfrac{1}{3.5}+\dfrac{1}{5.7}+...+\dfrac{1}{2007.2009}+\dfrac{1}{2009.2011}\)
\(=\dfrac{1}{2}\left(1-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+...+\dfrac{1}{2007}-\dfrac{1}{2009}+\dfrac{1}{2009}-\dfrac{1}{2011}\right)\)
\(=\dfrac{1}{2}\left(1-\dfrac{1}{2011}\right)\)
\(=\dfrac{1005}{2011}\)
NT
2
H9
HT.Phong (9A5)
CTVHS
12 tháng 8 2023
\(\left(2^3\cdot9^4+9^3\cdot45\right):\left(9^2\cdot10-9^2\right)\)
\(=\left(2^3\cdot3^8+3^6\cdot5\cdot3^2\right):\left[9^2\cdot\left(10-1\right)\right]\)
\(=\left(2^3\cdot3^8+3^8\cdot5\right):\left(9^2\cdot9\right)\)
\(=\left[3^8\cdot\left(2^3+5\right)\right]:9^3\)
\(=3^8\cdot13:3^6\)
\(=3^2\cdot13\)
\(=117\)
12 tháng 8 2023
\(\left(2^3.9^4+9^3.45\right):\left(9^2.10-9^2\right)\)
\(=\left(2^3.3^8+3^6.3^2.5\right):\left(3^4.10-3^4\right)\)
\(=\left(2^3.3^8+3^8.5\right):\left(3^4.\left(10-1\right)\right)\)
\(=3^8\left(2^3+5\right):3^4.9=3^8\left(2^3+5\right):3^6\)
\(=3^2.13=9.13=117\)
BF
26 tháng 2 2023
`(64×15×9)/(72×45×6) `
`=(8xx8×3xx5×9)/(8xx9×5xx9×2xx3) `
`=8/(9×2) `
`= 4/9`
KH
7
NT
Nguyễn Thị Thương Hoài
Giáo viên
VIP
28 tháng 5 2022
11/9 + 18/5 -2/9 -3/5
= 11/9 - 2/ 9 + 18/5 - 3/5
= 9/9 + 15/5
= 1+ 3
= 4
24 tháng 3 2021
\(A,\dfrac{7}{11}+\dfrac{5}{6}-\dfrac{-4}{11}-\dfrac{1}{6}\)
\(= (\dfrac{7}{11}+\dfrac{-4}{11})-(\dfrac{5}{6}-\dfrac{1}{6})\)
\(=\dfrac{3}{11}-\dfrac{2}{3}\)
\(= \dfrac{9}{33}-\dfrac{22}{33}\)
\(= \dfrac{-13}{33}\)
\(B,\dfrac{5}{9}.\dfrac{7}{3}+\dfrac{5}{9}.\dfrac{9}{13}-\dfrac{5}{9}:\dfrac{13}{5}\)
\(=(\dfrac{5}{9}.\dfrac{5}{9}:\dfrac{5}{9}).\dfrac{7}{3}+\dfrac{9}{13}-\dfrac{13}{5}\)
\(=(\dfrac{5}{9}.\dfrac{7}{3})+(\dfrac{9}{13}-\dfrac{13}{5})\)
\(=(\dfrac{5}{9}.\dfrac{21}{9})+(\dfrac{45}{65}-\dfrac{169}{65})\)
\(=\dfrac{35}{27}+\dfrac{-124}{65}\)
\(=\dfrac{2275}{1755}+\dfrac{-3348}{1755}\)
\(=\dfrac{-1073}{1755}\)
TT
21
QD
1
\(A=1+\dfrac{1}{1+2}+...+\dfrac{1}{1+2+...+8}\)
\(=\dfrac{2}{2}+\dfrac{1}{2\cdot\dfrac{3}{2}}+...+\dfrac{1}{8\cdot\dfrac{9}{2}}\)
\(=\dfrac{2}{1\cdot2}+\dfrac{2}{2\cdot3}+...+\dfrac{2}{8\cdot9}\)
\(=2\left(\dfrac{1}{1\cdot2}+\dfrac{1}{2\cdot3}+...+\dfrac{1}{8\cdot9}\right)\)
\(=2\left(1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{8}-\dfrac{1}{9}\right)\)
\(=2\left(1-\dfrac{1}{9}\right)=2\cdot\dfrac{8}{9}=\dfrac{16}{9}\)