\(\left(1-\frac{2}{6}\right)\times\left(1-\frac{2}{12}\right)\times\left(1-\frac{2}{20}\right)...\left(1-\frac{2}{9900}\right)\) Tính
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MT
19 tháng 3 2019
biết làm bài 1 thôi
\(\left(\frac{1}{2}+1\right)\times\left(\frac{1}{3}+1\right)\times\cdot\cdot\cdot\times\left(\frac{1}{999}+1\right)\)
= \(\frac{3}{2}\times\frac{4}{3}\times\frac{5}{4}\times\cdot\cdot\cdot\times\frac{1000}{999}\)
lượt bỏ đi còn :
\(\frac{1000}{2}=500\)
9 tháng 8 2017
a) \(\frac{\frac{2}{7}+\frac{2}{5}+\frac{2}{17}+\frac{2}{293}}{\frac{3}{7}+\frac{3}{5}+\frac{3}{17}+\frac{3}{293}}+\frac{\frac{7}{12}+\frac{5}{6}-1}{5-\frac{3}{4}+\frac{1}{3}}\) \(=\frac{2\left(\frac{1}{7}+\frac{1}{5}+\frac{1}{17}+\frac{1}{293}\right)}{3\left(\frac{1}{7}+\frac{1}{5}+\frac{1}{17}+\frac{1}{293}\right)}+\frac{\frac{5}{12}}{\frac{55}{12}}\)
\(=\frac{2}{3}+\frac{1}{11}=\frac{25}{33}\)
b) \(\left(1-\frac{1}{7}\right).\left(1-\frac{2}{7}\right)....\left(1-\frac{10}{7}\right)=\left(1-\frac{1}{7}\right).\left(1-\frac{2}{7}\right)...\left(1-\frac{7}{7}\right).\left(1-\frac{8}{7}\right).\left(1-\frac{9}{7}\right).\) \(\left(1-\frac{10}{7}\right)\) = 0
9 tháng 8 2017
a)\(\frac{\frac{2}{7}+\frac{2}{5}+\frac{2}{17}+\frac{2}{293}}{\frac{3}{7}+\frac{3}{5}+\frac{3}{17}+\frac{3}{293}}+\frac{\frac{7}{12}+\frac{5}{6}-1}{5-\frac{3}{4}+\frac{1}{3}}\)
\(=\frac{2\left(\frac{1}{7}+\frac{1}{5}+\frac{1}{17}+\frac{1}{293}\right)}{3\left(\frac{1}{7}+\frac{1}{5}+\frac{1}{17}+\frac{1}{293}\right)}+\frac{\frac{7}{12}+\frac{10}{12}-\frac{12}{12}}{\frac{60}{12}-\frac{9}{12}+\frac{4}{12}}\)
\(=\frac{2}{3}+\frac{\frac{5}{12}}{\frac{55}{12}}\)
\(=\frac{2}{3}+\frac{1}{11}\)
\(=\frac{25}{33}\)
b)\(\left(1-\frac{1}{7}\right)\cdot\left(1-\frac{2}{7}\right)\cdot...\cdot\left(1-\frac{10}{7}\right)\)
Ta nhận thấy trong tích này có 1 thừa số là\(\left(1-\frac{7}{7}\right)=0\)nên tích trên sẽ bằng 0.
TN
12 tháng 5 2016
\(=\frac{1}{2}\times\frac{2}{3}\times....\times\frac{2003}{2004}\)
\(=\frac{1\times2\times3\times...\times2003}{2\times3\times4\times...\times2014}\)
\(=\frac{1}{2014}\)
4 tháng 3 2018
\(A=\frac{1}{2}\left(1+\frac{1}{1\cdot3}\right)\left(1+\frac{1}{2\cdot4}\right)...\left(1+\frac{1}{2015\cdot2017}\right)\)\(A=\frac{1}{2}\left(\frac{1\cdot3+1}{1\cdot3}\right)\left(\frac{2\cdot4+1}{2\cdot4}\right)...\left(\frac{2015\cdot2017+1}{2015\cdot2017}\right)\)
\(A=\frac{1^2}{2}\cdot\frac{2^2}{1\cdot3}\cdot\frac{3^2}{2\cdot4}\cdot\cdot\cdot\frac{2016^2}{2015\cdot2017}\)
\(A=\frac{1^2\cdot2^2\cdot3^2\cdot\cdot\cdot2016^2}{2\cdot1\cdot3\cdot2\cdot4\cdot\cdot\cdot2015\cdot2017}\)
\(A=\frac{2016}{2017}\)
(1 - 2/6) × (1 - 2/12) × (1 - 2/20) × ... × (1 - 2/9900)
= 4/6 × 10/12 × 18/20 × ... × 9898/9900
= 1.4/2.3 × 2.5/3.4 × 3.6/4.5 × ... × 98.101/99.100
= 1.2.3...98/3.4.5...100 × 4.5.6...101/2.3.4...99
= 2/99.100 × 100.101/2.3
= 101/99×3 = 101/297
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