Bài 21:
\(A=\dfrac{1}{2}+\dfrac{1}{6}+\dfrac{1}{12}+...+\dfrac{1}{9900}\\ =\dfrac{1}{1\cdot2}+\dfrac{1}{2\cdot3}+\dfrac{1}{3\cdot4}+..+\dfrac{1}{99\cdot100}\\ =1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{99}-\dfrac{1}{100}\\ =1-\dfrac{1}{100}\\ =\dfrac{99}{100}\\ B=\dfrac{2}{3}+\dfrac{2}{15}+\dfrac{2}{35}+...+\dfrac{2}{9999}\\ =\dfrac{2}{1\cdot3}+\dfrac{2}{3\cdot5}+\dfrac{2}{5\cdot7}+...+\dfrac{2}{99\cdot101}\\ =1-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+...+\dfrac{1}{99}-\dfrac{1}{101}\\ =1-\dfrac{1}{101}\\ =\dfrac{100}{101}\\ C=\dfrac{5}{1\cdot2}+\dfrac{5}{2\cdot3}+...+\dfrac{5}{99\cdot100}\\ =5\left(\dfrac{1}{1\cdot2}+\dfrac{1}{2\cdot3}+..+\dfrac{1}{99\cdot100}\right)\\ =5\left(1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{99}-\dfrac{1}{100}\right)\\ =5\left(1-\dfrac{1}{100}\right)\\ =5\cdot\dfrac{99}{100}\\ =\dfrac{99}{20}\\ D=\dfrac{5}{1\cdot3}+\dfrac{5}{3\cdot5}+...+\dfrac{5}{99\cdot101}\\ =\dfrac{5}{2}\cdot\left(\dfrac{2}{1\cdot3}+\dfrac{2}{3\cdot5}+...+\dfrac{2}{99\cdot101}\right)\\ =\dfrac{5}{2}\cdot\left(1-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+...+\dfrac{1}{99}-\dfrac{1}{101}\right)\\ =\dfrac{5}{2}\left(1-\dfrac{1}{101}\right)\\ =\dfrac{5}{2}\cdot\dfrac{100}{101}=\dfrac{250}{101}\)
Bài 20:
a: \(\dfrac{7}{25}+\dfrac{2}{5}+\dfrac{18}{25}+\dfrac{3}{5}+\dfrac{1}{2}\)
\(=\left(\dfrac{7}{25}+\dfrac{18}{25}\right)+\left(\dfrac{2}{5}+\dfrac{3}{5}\right)+\dfrac{1}{2}\)
\(=1+1+\dfrac{1}{2}=\dfrac{5}{2}\)
b: \(\dfrac{16}{5}\cdot\dfrac{7}{11}+\dfrac{4}{11}\cdot\dfrac{16}{5}-\dfrac{6}{5}\)
\(=\dfrac{16}{5}\left(\dfrac{7}{11}+\dfrac{4}{11}\right)-\dfrac{6}{5}\)
\(=\dfrac{16}{5}-\dfrac{6}{5}=\dfrac{10}{5}=2\)
c: \(3\dfrac{5}{7}-2\dfrac{3}{5}+4\dfrac{2}{7}\)
\(=\left(3-2+4\right)+\left(\dfrac{5}{7}-\dfrac{3}{5}+\dfrac{2}{7}\right)\)
\(=5+1-\dfrac{3}{5}=6-\dfrac{3}{5}=\dfrac{27}{5}\)
d: \(\dfrac{\dfrac{1}{9}-\dfrac{1}{7}-\dfrac{1}{11}}{\dfrac{4}{9}-\dfrac{4}{7}-\dfrac{4}{11}}=\dfrac{\dfrac{1}{9}-\dfrac{1}{7}-\dfrac{1}{11}}{4\left(\dfrac{1}{9}-\dfrac{1}{7}-\dfrac{1}{11}\right)}=\dfrac{1}{4}\)
e: \(\dfrac{2022\cdot2023-145}{1877+2022\cdot2022}\)
\(=\dfrac{2022\cdot\left(2022+1\right)-145}{1877+2022^2}\)
\(=\dfrac{2022^2+2022-145}{1877+2022^2}=1\)
Bài 21:
a: \(A=\dfrac{1}{2}+\dfrac{1}{6}+...+\dfrac{1}{9900}\)
\(=\dfrac{1}{1\cdot2}+\dfrac{1}{2\cdot3}+...+\dfrac{1}{99\cdot100}\)
\(=1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{99}-\dfrac{1}{100}\)
\(=1-\dfrac{1}{100}=\dfrac{99}{100}\)
b: \(B=\dfrac{2}{3}+\dfrac{2}{15}+...+\dfrac{2}{9999}\)
\(=\dfrac{2}{1\cdot3}+\dfrac{2}{3\cdot5}+...+\dfrac{2}{99\cdot101}\)
\(=1-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+...+\dfrac{1}{99}-\dfrac{1}{101}=1-\dfrac{1}{101}=\dfrac{100}{101}\)
c: \(C=\dfrac{5}{1\cdot2}+\dfrac{5}{2\cdot3}+...+\dfrac{5}{99\cdot100}\)
\(=5\left(\dfrac{1}{1\cdot2}+\dfrac{1}{2\cdot3}+...+\dfrac{1}{99\cdot100}\right)\)
\(=5\left(1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{99}-\dfrac{1}{100}\right)\)
\(=5\left(1-\dfrac{1}{100}\right)=5\cdot\dfrac{99}{100}=\dfrac{99}{20}\)
d: \(D=\dfrac{5}{1\cdot3}+\dfrac{5}{3\cdot5}+...+\dfrac{5}{99\cdot101}\)
\(=\dfrac{5}{2}\left(\dfrac{2}{1\cdot3}+\dfrac{2}{3\cdot5}+...+\dfrac{2}{99\cdot101}\right)\)
\(=\dfrac{5}{2}\left(1-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+...+\dfrac{1}{99}-\dfrac{1}{101}\right)=\dfrac{5}{2}\left(1-\dfrac{1}{101}\right)\)
\(=\dfrac{5}{2}\cdot\dfrac{100}{101}=\dfrac{500}{202}=\dfrac{250}{101}\)