a, \(\dfrac{2}{1.3}+\dfrac{2}{3.5}+\dfrac{2}{5.7}+...+\dfrac{2}{99.101}\\ =1-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+...+\dfrac{1}{99}-\dfrac{1}{101}\\ =1-\dfrac{1}{101}=\dfrac{100}{101}\)
b, \(\dfrac{5}{1.3}+\dfrac{5}{3.5}+...+\dfrac{5}{99.101}\)
\(=\dfrac{5}{2}\left(\dfrac{2}{1.3}+\dfrac{2}{3.5}+...+\dfrac{2}{99.101}\right)\)
\(=\dfrac{5}{2}\left(1-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+...+\dfrac{1}{99}-\dfrac{1}{101}\right)\)
\(=\dfrac{5}{2}\left(1-\dfrac{1}{101}\right)\)
\(=\dfrac{5}{2}.\dfrac{100}{101}=\dfrac{250}{101}\)
Vậy...
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+\(=1-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+...+\dfrac{1}{99}-\dfrac{1}{101}\)
\(=1-\dfrac{1}{101}\)
\(=\dfrac{101}{101}+\dfrac{-1}{101}\)
\(=\dfrac{100}{101}\)
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Gọi tổng của các phân số trên là A.
Ta có:
\(A=\dfrac{2}{2}.\left(\dfrac{1}{1}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+...+\dfrac{1}{99}-\dfrac{1}{101}\right)\)
\(A=1.\left(\dfrac{1}{1}-\dfrac{1}{101}\right)\)
\(A=1.\dfrac{100}{101}=\dfrac{100}{101}\)
Vậy \(A=\dfrac{100}{101}.\)
b)
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Gọi tổng các phân số trên là B.
Ta có:
\(B=\dfrac{5}{2}.\left(\dfrac{1}{1}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+...+\dfrac{1}{99}-\dfrac{1}{101}\right)\)
\(B=\dfrac{5}{2}.\left(\dfrac{1}{1}-\dfrac{1}{101}\right)\)
\(B=\dfrac{5}{2}.\dfrac{100}{101}\)
\(B=\dfrac{250}{101}\)
Vậy \(B=\dfrac{250}{101}\).