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25 tháng 7 2018

\(\left(1+\dfrac{1}{2.4}\right)\left(1+\dfrac{1}{3.5}\right)\left(1+\dfrac{1}{4.6}\right)...\left(1+\dfrac{1}{99.101}\right)\)

\(=\dfrac{3^2}{2.4}.\dfrac{4^2}{3.5}.\dfrac{5^2}{4.6}...\dfrac{100^2}{99.101}\)

\(=\dfrac{3^2.4^2.5^2...100^2}{2.3.4^2.5^2...99^2.100.101}\)

\(=\dfrac{3^2.100^2}{2.3.100.101}\)

\(=\dfrac{3.100}{2.101}\)

\(=\dfrac{300}{202}\)

25 tháng 7 2018

Tớ nghĩ đề là :

\(\left(1+\dfrac{1}{2.4}\right)\left(1+\dfrac{1}{3.5}\right)\left(1+\dfrac{1}{4.6}\right)...\left(1+\dfrac{1}{99.101}\right)\)

\(=\dfrac{9}{2.4}.\dfrac{16}{3.5}.\dfrac{25}{4.6}....\dfrac{10000}{99.101}=\dfrac{3.3.4.4.5...100.100}{2.4.3.5.4....99.101}\)

\(=\dfrac{3.100}{2.101}=\dfrac{150}{101}\)

10 tháng 4 2018

\(\left(1+\frac{1}{1\cdot3}\right)\left(1+\frac{1}{2\cdot4}\right)\left(1+\frac{1}{3\cdot5}\right)...\left(1+\frac{1}{2013\cdot2015}\right)\)

\(=\frac{4}{1\cdot3}\cdot\frac{9}{2\cdot4}\cdot\frac{16}{3\cdot5}\cdot...\cdot\frac{4056196}{2013\cdot2015}\)

\(=\frac{\left(2\cdot2\right)\left(3\cdot3\right)\left(4\cdot4\right)...\left(2014\cdot2014\right)}{\left(1\cdot3\right)\left(2\cdot4\right)\left(3\cdot5\right)...\left(2013\cdot2015\right)}\)

\(=\frac{\left(2\cdot3\cdot4\cdot...\cdot2014\right)\left(2\cdot3\cdot4\cdot...\cdot2014\right)}{\left(1\cdot2\cdot3\cdot...\cdot2013\right)\left(3\cdot4\cdot5\cdot...\cdot2015\right)}\)

\(=\frac{2014\cdot2}{1\cdot2015}\)

\(=\frac{4028}{2015}\)

11 tháng 4 2018

\(\left(1+\frac{1}{1.3}\right)\left(1+\frac{1}{2.4}\right)...\left(1+\frac{1}{99.101}\right)\)

\(=\frac{4}{1.3}.\frac{9}{2.4}....\frac{10000}{99.101}\)

\(=\frac{2.2.3.3...100.100}{1.3.2.4...99.101}\)

\(=\frac{\left(2.3.4...100\right)\left(2.3.4...100\right)}{\left(1.2...99\right)\left(3.4.5...101\right)}\)

\(=\frac{100.2}{101}=\frac{200}{101}\)

12 tháng 4 2021

\(D=\left(1+\frac{1}{1.3}\right)\left(1+\frac{1}{2.4}\right)\left(1+\frac{1}{3.5}\right)...\left(1+\frac{1}{99.101}\right)\)

\(D=\frac{4}{1.3}.\frac{9}{2.4}.\frac{16}{3.5}...\frac{10000}{99.101}\)

\(D=\frac{2^2}{1.3}.\frac{3^2}{2.4}.\frac{4^2}{3.5}...\frac{100^2}{99.101}\)

\(D=\frac{2.3.4...100}{1.2.3...99}.\frac{2.3.4...100}{3.4.5...101}=100.\frac{2}{101}=\frac{200}{101}\)

Vậy  \(D=\frac{200}{101}\)