tinh nhanh\(\left(1-\dfrac{1}{21}\right).\left(1-\dfrac{1}{28}\right)....\left(1-\dfrac{1}{1326}\right)\)
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18 tháng 3 2017
a)
\(A=\dfrac{3}{4}.\dfrac{8}{9}...\dfrac{9999}{10000}\)
\(=\dfrac{1.3}{2.2}.\dfrac{2.4}{3.3}...\dfrac{99.101}{100.100}\)
\(=\dfrac{1.2...99}{2.3...100}.\dfrac{3.4...101}{2.3...100}\)
\(=\dfrac{1}{100}.\dfrac{101}{2}\)
\(=\dfrac{101}{200}\)
NV
Nguyễn Việt Lâm
Giáo viên
24 tháng 2 2019
\(\left(1-\dfrac{28}{10}\right)\left(1-\dfrac{52}{22}\right)\left(1-\dfrac{80}{36}\right)...\left(1-\dfrac{21808}{10900}\right)\)
\(=\left(1-\dfrac{2.10+8}{10}\right)\left(1-\dfrac{2.22+8}{22}\right)\left(1-\dfrac{2.36+8}{36}\right)...\left(1-\dfrac{2.10900+8}{10900}\right)\)
\(=\left(1-2-\dfrac{8}{10}\right)\left(1-2-\dfrac{8}{22}\right)\left(1-2-\dfrac{8}{36}\right)...\left(1-2-\dfrac{8}{10900}\right)\)
\(=\left(-1-\dfrac{8}{1.10}\right)\left(-1-\dfrac{8}{2.11}\right)\left(-1-\dfrac{8}{3.12}\right)...\left(-1-\dfrac{8}{100.109}\right)\)
\(=\left(\dfrac{-18}{1.10}\right)\left(\dfrac{-30}{2.11}\right)\left(\dfrac{-44}{3.12}\right)...\left(\dfrac{-10908}{100.109}\right)\)
\(=\left(\dfrac{-2.9}{1.10}\right)\left(\dfrac{-3.10}{2.11}\right)\left(\dfrac{-4.11}{3.12}\right)...\left(\dfrac{-101.108}{100.109}\right)\)
\(=\dfrac{\left(-2\right)\left(-3\right)\left(-4\right)...\left(-101\right)}{1.2.3...100}.\dfrac{9.10.11...108}{10.11.12...109}\) (1)
\(=\dfrac{101}{1}.\dfrac{9}{109}=\dfrac{909}{109}\)
Do ở (1) có \(-2-\left(-101\right)+1=100\) nhân tử (số nhân tử là số chẵn) mang dấu âm nên kết quả sẽ mang dấu dương
PT
23 tháng 9 2017
\(A=\left(1-\dfrac{1}{15}\right).\left(1-\dfrac{1}{21}\right).\left(1-\dfrac{1}{28}\right)....\left(1-\dfrac{1}{210}\right)\)
\(A=\dfrac{14}{15}.\dfrac{20}{21}.\dfrac{27}{28}....\dfrac{209}{210}\)
\(A=\dfrac{14.2}{15.2}.\dfrac{20.2}{21.2}.\dfrac{27.2}{28.2}....\dfrac{209.2}{210.2}\)
\(A=\dfrac{4.7}{5.6}.\dfrac{5.8}{6.7}.\dfrac{6.9}{7.8}....\dfrac{19.22}{20.21}\)
\(A=\dfrac{4.5.6....19}{5.6.7....20}.\dfrac{7.8.9....22}{6.7.8....21}\)
\(A=\dfrac{4}{20}.\dfrac{22}{6}\)
\(A=\dfrac{11}{15}\)
19 tháng 11 2022
a: \(=\left(\dfrac{15}{34}+\dfrac{19}{34}\right)+\left(\dfrac{7}{21}+\dfrac{2}{3}\right)-\dfrac{32}{17}=2-\dfrac{32}{17}=\dfrac{2}{17}\)
b: \(=-8\cdot\dfrac{1}{4}:\left(\dfrac{9}{4}-\dfrac{7}{6}\right)=-2:\dfrac{27-14}{12}=\dfrac{-2\cdot12}{13}=-\dfrac{24}{13}\)
c: \(=\dfrac{-5}{3}\left(16+\dfrac{2}{7}+28+\dfrac{2}{7}\right)=\dfrac{-5}{3}\cdot\left(44+\dfrac{4}{7}\right)\)
=-520/7
24 tháng 8 2017
mk chỉ giải đc câu d thôi nha ; bn thông cảm
d) \(D=\dfrac{2006.2005-1}{2004.2006+2005}=\dfrac{2006.2005-1}{2004.2006+2006-1}\)
\(D=\dfrac{2006.2005-1}{2006\left(2004+1\right)-1}=\dfrac{2006.2005-1}{2006.2005-1}=1\)
MN
11 tháng 11 2018
e)\(16\dfrac{2}{7}:\left(-\dfrac{3}{5}\right)+28\dfrac{2}{7}:\left(-\dfrac{3}{5}\right)\)
=\(\left(16\dfrac{2}{7}+28\dfrac{2}{7}\right):\left(-\dfrac{3}{5}\right)\)
=\(\dfrac{312}{7}\)\(:\left(-\dfrac{3}{5}\right)\)
=\(-\dfrac{516}{7}\)
MN
11 tháng 11 2018
a)\(\dfrac{7}{8}.\left(\dfrac{2}{12}+\dfrac{4}{10}\right)\)
=\(\dfrac{7}{8}.\left(\dfrac{1}{6}+\dfrac{2}{5}\right)\)
=\(\dfrac{7}{8}.\)\(\dfrac{17}{30}\)
=\(\dfrac{119}{240}\)
14 tháng 3 2017
Ta có: \(A=\left(\dfrac{1}{3}-1\right)\left(\dfrac{1}{6}-1\right)\left(\dfrac{1}{10}-1\right)\left(\dfrac{1}{15}-1\right)\left(\dfrac{1}{21}-1\right)\left(\dfrac{1}{28}-1\right)\left(\dfrac{1}{36}-1\right)\)
\(=\dfrac{-2}{3}.\dfrac{-5}{6}.\dfrac{-9}{10}.\dfrac{-14}{15}.\dfrac{-20}{21}.\dfrac{-27}{28}.\dfrac{-35}{36}\)
\(=\dfrac{-2.\left(-5\right).3.\left(-3\right).2.\left(-7\right).\left(-4\right).5.\left(-3\right).9.5.\left(-7\right)}{3.2.3.2.5.3.5.3.7.4.7.4.9}\)
\(=\dfrac{-5}{3.4}=\dfrac{-5}{12}\)
Vậy \(A=\dfrac{-5}{12}.\)
AT
14 tháng 3 2017
\(C=1+\dfrac{1}{2}+\dfrac{1}{2^2}+\dfrac{1}{2^3}+...+\dfrac{1}{2^{2016}}\)
\(2C=2\left(1+\dfrac{1}{2}+\dfrac{1}{2^2}+\dfrac{1}{2^3}+...+\dfrac{1}{2^{2016}}\right)\)
\(2C=2+1+\dfrac{1}{2}+\dfrac{1}{2^2}+....+\dfrac{1}{2^{2015}}\)
\(2C-C=\left(2+1+\dfrac{1}{2}+...+\dfrac{1}{2^{2015}}\right)-\left(1+\dfrac{1}{2}+\dfrac{1}{2^2}+...+\dfrac{1}{2^{2016}}\right)\)
\(C=2-\dfrac{1}{2^{2016}}\)
29 tháng 3
a)\(\dfrac{-10}{11}.\dfrac{8}{9}+\dfrac{7}{18}.\dfrac{10}{11}\)
=\(\dfrac{10}{11}.\dfrac{-8}{9}+\dfrac{7}{18}.\dfrac{10}{11}\)
=\(\dfrac{10}{11}(\dfrac{-8}{9}+\dfrac{7}{18})\)
=\(\dfrac{10}{11}.\dfrac{-1}{2}\)
=\(\dfrac{-5}{11}\)
NT
Nguyễn Thị Thương Hoài
Giáo viên
VIP
7 tháng 8
b;
B = \(\dfrac{3}{14}\) : \(\dfrac{1}{28}\) - \(\dfrac{13}{21}\): \(\dfrac{1}{28}\) + \(\dfrac{29}{42}\) : \(\dfrac{1}{28}\) - 8
B = (\(\dfrac{3}{14}\) - \(\dfrac{13}{21}\) + \(\dfrac{29}{42}\)) - 8
B = (\(\dfrac{9}{42}\) - \(\dfrac{26}{42}\) + \(\dfrac{29}{42}\)) - 8
B = (\(\dfrac{-17}{42}\) + \(\dfrac{29}{42}\)) - 8
B = \(\dfrac{2}{7}\) - 8
B = \(\dfrac{2}{7}-\dfrac{56}{7}\)
B = - \(\dfrac{54}{7}\)
\(\left(1-\dfrac{1}{21}\right)\left(1-\dfrac{1}{28}\right)...\left(1-\dfrac{1}{1326}\right)\)
\(=\left(1-\dfrac{2}{42}\right)\left(1-\dfrac{2}{56}\right)\left(1-\dfrac{2}{72}\right)...\left(1-\dfrac{2}{2652}\right)\)
\(=\dfrac{40}{42}.\dfrac{54}{56}.\dfrac{70}{72}...\dfrac{2650}{2652}\)
\(=\dfrac{5.8}{6.7}.\dfrac{6.9}{7.8}.\dfrac{7.10}{8.9}...\dfrac{50.53}{51.52}\)
\(=\dfrac{5.6.7...50}{6.7.8...51}.\dfrac{8.9.10...53}{7.8.9...52}\)
\(=\dfrac{5}{51}.\dfrac{53}{7}=\dfrac{265}{357}\)