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\(\left(1+\frac{1}{100}\right).\left(1+\frac{1}{99}\right)............\left(1+\frac{1}{2}\right)\)
\(=\frac{101}{100}.\frac{100}{99}..............\frac{3}{2}\)
\(=\frac{101.100............3}{100.99...............2}\)
\(=\frac{101}{2}\)
\(=\frac{101}{100}.\frac{100}{989}.....\frac{3}{2}=\frac{101}{2}\)
1/ Đặt \(A=1+\frac{1}{3}+\frac{1}{9}+\frac{1}{27}+\frac{1}{81}+\frac{1}{243}\)
=> \(A=1+\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+\frac{1}{3^4}+\frac{1}{3^5}\) => \(3A=3+1+\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+\frac{1}{3^4}\)
=> \(3A-A=3+1+\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+\frac{1}{3^4}-\left(1+\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+\frac{1}{3^4}+\frac{1}{3^5}\right)\)
=> \(2A=3-\frac{1}{3^5}\) => \(A=\frac{3}{2}-\frac{1}{2.3^5}\)
Tổ 1 chiếm số phần là: \(\frac{18}{90}=\frac{1}{5}\)(Tổng số công nhân)
Tổng số phần của tổ 1 và tổ 2 chiếm số phần là: \(\frac{1}{5}+\frac{3}{10}=\frac{5}{10}=\frac{1}{2}\)(Tổng số công nhân)
=> Tổ 3 chiếm số phần là: \(1-\frac{1}{2}=\frac{1}{2}\)(Tổng số công nhân)
Đáp số: 1/2 (Tổng số công nhân)
\(\frac{2}{3}:\left(\frac{2}{1}-\frac{2}{3}-\frac{2}{5}-\frac{2}{7}-\frac{2}{9}-\frac{2}{11}\right)x\frac{1}{2}\)
\(=\frac{211}{1155}\)
\(\frac{1}{7}\times3\frac{1}{11}-\frac{1}{7}\times\frac{9}{11}-\frac{1}{7}\)
\(=\frac{1}{7}\times\frac{34}{11}-\frac{1}{7}\times\frac{9}{11}-\frac{1}{7}\)
\(=\frac{1}{7}\left(\frac{34}{11}-\frac{9}{11}-\frac{11}{11}\right)\)
\(=\frac{1}{7}\times\frac{14}{11}\)
\(=\frac{2}{11}\)
\(1\frac{1}{3}+1\frac{1}{5}.y-\frac{4}{5}=2\frac{4}{5}\)
\(\Rightarrow\frac{4}{3}+\frac{6}{5}.y=2\frac{4}{5}+\frac{4}{5}\)
\(\Rightarrow\frac{4}{3}+\frac{6}{5}.y=2\frac{8}{5}=\frac{18}{5}\)
\(\Rightarrow\frac{6}{5}y=\frac{18}{5}-\frac{4}{3}\)
\(\Rightarrow\frac{6}{5}y=\frac{34}{15}\)
\(\Rightarrow y=\frac{34}{15}:\frac{6}{5}\)
\(\Rightarrow y=\frac{34}{15}.\frac{5}{6}=\frac{17}{9}\)
\(\frac{1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{99}}{\frac{1}{99}+\frac{2}{98}+\frac{3}{97}+...+\frac{99}{1}}\)
\(=\frac{1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{99}}{\left(1+\frac{1}{99}\right)+\left(1+\frac{2}{98}\right)++...+\left(1+\frac{98}{2}\right)1}\)
\(=\frac{1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{99}}{\frac{100}{99}+\frac{100}{98}+...+\frac{100}{2}+\frac{100}{100}}\)
\(=\frac{1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{99}}{100\times\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{99}\right)}\)
\(=\frac{1}{100}\)
=\(\frac{101}{2}\)
cac so triet tieu cho nhau roi \(=\frac{101}{100}\cdot\frac{100}{99}\cdot\frac{99}{98}\cdot.....\cdot\frac{4}{3}\cdot\frac{3}{2}\)
cac so tu 3 den 100 bi triet tieu
=\(\frac{101}{2}\)
\(\left(1+\frac{1}{100}\right).\left(1+\frac{1}{99}\right)...\left(1+\frac{1}{3}\right).\left(1+\frac{1}{2}\right)\)
\(=\frac{101}{100}.\frac{100}{99}...\frac{4}{3}.\frac{3}{2}\)
\(=\frac{101.\left(100...4.3\right)}{\left(100.99...3\right).2}\)
\(=\frac{101}{2}\)