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\(\left(1+\frac{1}{1.3}\right)\left(1+\frac{1}{2.4}\right)...\left(1+\frac{1}{99.101}\right)\)
\(=\frac{4}{1.3}.\frac{9}{2.4}....\frac{10000}{99.101}\)
\(=\frac{2.2.3.3...100.100}{1.3.2.4...99.101}\)
\(=\frac{\left(2.3.4...100\right)\left(2.3.4...100\right)}{\left(1.2...99\right)\left(3.4.5...101\right)}\)
\(=\frac{100.2}{101}=\frac{200}{101}\)
\(D=\left(1+\frac{1}{1.3}\right)\left(1+\frac{1}{2.4}\right)\left(1+\frac{1}{3.5}\right)...\left(1+\frac{1}{99.101}\right)\)
\(D=\frac{4}{1.3}.\frac{9}{2.4}.\frac{16}{3.5}...\frac{10000}{99.101}\)
\(D=\frac{2^2}{1.3}.\frac{3^2}{2.4}.\frac{4^2}{3.5}...\frac{100^2}{99.101}\)
\(D=\frac{2.3.4...100}{1.2.3...99}.\frac{2.3.4...100}{3.4.5...101}=100.\frac{2}{101}=\frac{200}{101}\)
Vậy \(D=\frac{200}{101}\)
\(\left(1+\frac{1}{1.3}\right).\left(1+\frac{1}{2.4}\right).\left(\frac{1}{3.5.}\right).....\left(1+\frac{1}{99.101}\right)\)
\(=\frac{4}{3}.\frac{9}{8}.\frac{16}{15}.....\frac{10000}{9999}\)
\(=\frac{2^2}{1.3}.\frac{3^2}{2.4}.\frac{4^2}{3.5}.....\frac{100^2}{99.101}\)
\(=\frac{2^2.3^2.4^2.5^2.....98^2.99^2.100^2}{1.2.3^2.4^2.5^2......99^2.100.101}\)
\(=\frac{2.100}{1.101}\)
\(=\frac{200}{101}\)
Tớ nghĩ đề là :
\(\left(1+\dfrac{1}{2.4}\right)\left(1+\dfrac{1}{3.5}\right)\left(1+\dfrac{1}{4.6}\right)...\left(1+\dfrac{1}{99.101}\right)\)
\(=\dfrac{9}{2.4}.\dfrac{16}{3.5}.\dfrac{25}{4.6}....\dfrac{10000}{99.101}=\dfrac{3.3.4.4.5...100.100}{2.4.3.5.4....99.101}\)
\(=\dfrac{3.100}{2.101}=\dfrac{150}{101}\)
\(\left(1+\dfrac{1}{2.4}\right)\left(1+\dfrac{1}{3.5}\right)\left(1+\dfrac{1}{4.6}\right)...\left(1+\dfrac{1}{99.101}\right)\)
\(=\dfrac{3^2}{2.4}.\dfrac{4^2}{3.5}.\dfrac{5^2}{4.6}...\dfrac{100^2}{99.101}\)
\(=\dfrac{3^2.4^2.5^2...100^2}{2.3.4^2.5^2...99^2.100.101}\)
\(=\dfrac{3^2.100^2}{2.3.100.101}\)
\(=\dfrac{3.100}{2.101}\)
\(=\dfrac{300}{202}\)