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\(\frac{1}{1\cdot3}+\frac{1}{3\cdot5}+\frac{1}{5\cdot7}+...+\frac{1}{99\cdot101}\)
\(=2\cdot\left(\frac{1}{1\cdot3}+\frac{1}{3\cdot5}+\frac{1}{5\cdot7}+...+\frac{1}{99\cdot101}\right):2\)
\(=\left(\frac{2}{1\cdot3}+\frac{2}{3\cdot5}+\frac{2}{5\cdot7}+...+\frac{2}{99\cdot101}\right):2\)
\(=\left(\frac{3-1}{1\cdot3}+\frac{5-3}{3\cdot5}+\frac{7-5}{5\cdot7}+...+\frac{101-99}{99\cdot101}\right):2\)
\(=\left(\frac{1}{1}-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{99}-\frac{1}{101}\right):2\)
\(=\left(\frac{1}{1}-\frac{1}{101}\right):2\)
\(=\frac{100}{101}:2=\frac{50}{101}\).
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\(\left(1+\frac{1}{1\times3}\right)\times\left(1+\frac{1}{2\times4}\right)\times\left(1+\frac{1}{3\times5}\right)\times...\times\left(1+\frac{1}{99.101}\right)\)
\(=\left(\frac{3}{3}+\frac{1}{3}\right)\times\left(\frac{8}{8}+\frac{1}{8}\right)\times\left(\frac{15}{15}+\frac{1}{15}\right)\times...\times\left(\frac{9999}{9999}+\frac{1}{9999}\right)\)
\(=\frac{4}{3}\times\frac{9}{8}\times\frac{16}{15}\times...\times\frac{10000}{9999}\)
\(=\frac{4\times9\times16\times...\times10000}{3\times8\times15\times...\times9999}\)
\(=\frac{2\times2\times3\times3\times4\times4\times...\times100\times100}{1\times3\times2\times4\times3\times5\times...\times99\times101}\)
\(=\frac{2\times100}{101}=\frac{200}{101}\)
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a, \(\frac{1}{5.6}+\frac{1}{6.7}+...+\frac{1}{49.50}\)
\(=\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7}+...+\frac{1}{49}-\frac{1}{50}\)
\(=\frac{1}{5}-\frac{1}{50}=\frac{9}{50}\)
b, \(\frac{2}{1.3}+\frac{2}{3.5}+...+\frac{2}{99.101}\)
\(=1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+...+\frac{1}{99}-\frac{1}{101}\)
\(=1-\frac{1}{101}=\frac{100}{101}\)
\(\frac{1}{5\times6}+\frac{1}{6\times7}+\frac{1}{7\times8}+...+\frac{1}{49\times50}\)
\(=\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7}+\frac{1}{7}-\frac{1}{8}+...+\frac{1}{49}-\frac{1}{50}\)
\(=\frac{1}{5}-\frac{1}{50}=\frac{9}{50}\)
~ Hok tốt ~
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Sửa đề: \(\dfrac{4}{1.3}+\dfrac{4}{3.5}+...+\dfrac{4}{99.101}\)
Đặt: \(A=\dfrac{4}{1.3}+\dfrac{4}{3.5}+...+\dfrac{4}{99.101}\)
\(=2\left(\dfrac{2}{1.3}+\dfrac{2}{3.5}+...+\dfrac{2}{99.101}\right)\)
\(=2\left(1-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+...+\dfrac{1}{97}-\dfrac{1}{99}+\dfrac{1}{99}-\dfrac{1}{101}\right)\)
\(=2\left(1-\dfrac{1}{101}\right)=\dfrac{200}{101}\)
*Lưu ý: Dấu ".'' trong bài là dấu nhân nhé, lên lớp 6 bạn sẽ được học
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Ta có : \(\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+.....+\frac{2}{99.101}\)
\(=1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+......+\frac{1}{99}-\frac{1}{101}\)
\(=1-\frac{1}{101}\)
\(=\frac{100}{101}\)
Đặt : \(A=\frac{2}{1\cdot3}+\frac{2}{3\cdot5}+\frac{2}{5\cdot7}+...+\frac{2}{99\cdot101}\)
\(A-\frac{2}{1\cdot3}=\frac{2}{3\cdot5}+\frac{2}{5\cdot7}+...+\frac{2}{99\cdot101}\)
\(2A-\frac{2}{1\cdot3}=\frac{2}{3}-\frac{2}{5}+\frac{2}{5}-\frac{2}{7}+\frac{2}{7}-...+\frac{2}{99}-\frac{2}{101}\)
\(2A-\frac{2}{3}=\frac{2}{3}-\frac{2}{101}\)
\(2A-\frac{2}{3}=\frac{196}{303}\)
\(A-\frac{2}{3}=\frac{98}{303}\)
\(A=\frac{98}{303}+\frac{2}{3}=\frac{100}{101}\)