giúp m9nhf bài này nhé
a: Ta có: \(\dfrac{1}{1\cdot2}+\dfrac{1}{2\cdot3}+\dfrac{1}{3\cdot4}+...+\dfrac{1}{2009\cdot2010}\)
\(=1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{2009}-\dfrac{1}{2010}\)
\(=1-\dfrac{1}{2010}=\dfrac{2009}{2010}\)
d) Đặt biểu thức trên là A
\(A=\dfrac{-1}{3}+\dfrac{-1}{15}+...+\dfrac{-1}{9999}\)
\(A=-\left(\dfrac{1}{3}+\dfrac{1}{15}+...+\dfrac{1}{9999}\right)\)
\(A=-\left(\dfrac{1}{1.3}+\dfrac{1}{3.5}+...+\dfrac{1}{99.101}\right)\)
\(2A=-\left(\dfrac{2}{1.3}+\dfrac{2}{3.5}+...+\dfrac{2}{99.101}\right)\)
\(2A=-\left(1-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+...+\dfrac{1}{99}-\dfrac{1}{101}\right)\)
\(2A=-\left(1-\dfrac{1}{101}\right)\)
\(2A=-\dfrac{100}{101}\Rightarrow A=\dfrac{-50}{101}\)
\(\dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+...+\dfrac{1}{2009.2010}\)
\(=1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{2009}-\dfrac{1}{2010}\)
\(=1-\dfrac{1}{2010}=\dfrac{2009}{2010}\)
\(\dfrac{1}{1.4}+\dfrac{1}{4.7}+\dfrac{1}{7.10}+...+\dfrac{1}{100.103}\)
\(3A=\dfrac{3}{1.4}+\dfrac{3}{4.7}+\dfrac{3}{7.10}+...+\dfrac{3}{100.103}\)
\(=1-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{10}+...+\dfrac{1}{100}-\dfrac{1}{103}\)
\(=1-\dfrac{1}{103}=\dfrac{102}{103}\)
\(\Rightarrow A=\dfrac{102}{309}=\dfrac{34}{103}\)
\(-\dfrac{1}{2011.2010}-\dfrac{1}{2010.2009}-...-\dfrac{1}{3.2}-\dfrac{1}{2.1}\)
\(=-\left(\dfrac{1}{1.2}+\dfrac{1}{2.3}+....+\dfrac{1}{2009.2010}+\dfrac{1}{2010.2011}\right)\)
\(=-\left(1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{2009}-\dfrac{1}{2010}+\dfrac{1}{2010}-\dfrac{1}{2011}\right)\)
\(=-\left(1-\dfrac{1}{2011}\right)=-\dfrac{2010}{2011}\)
\(\dfrac{-1}{3}+\dfrac{-1}{15}+\dfrac{-1}{35}+\dfrac{-1}{63}+...+\dfrac{-1}{9999}\)
\(=-\left(\dfrac{1}{1.3}+\dfrac{1}{3.5}+\dfrac{1}{5.7}+\dfrac{1}{7.9}+...+\dfrac{1}{9999}\right)\)
câu này b thông cảm mình kh tách được 9999 thành tích của 2 số lẻ liên tiếp nhau
b: Ta có: \(\dfrac{1}{1\cdot4}+\dfrac{1}{4\cdot7}+\dfrac{1}{7\cdot10}+...+\dfrac{1}{100\cdot103}\)
\(=\dfrac{1}{3}\left(\dfrac{3}{1\cdot4}+\dfrac{3}{4\cdot7}+...+\dfrac{3}{100\cdot103}\right)\)
\(=\dfrac{1}{3}\left(1-\dfrac{1}{103}\right)\)
\(=\dfrac{1}{3}\cdot\dfrac{102}{103}=\dfrac{34}{103}\)
c: Ta có: \(-\dfrac{1}{1\cdot2}-\dfrac{1}{2\cdot3}-...-\dfrac{1}{2009\cdot2010}-\dfrac{1}{2010\cdot2011}\)
\(=-\left(1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{2010}-\dfrac{1}{2011}\right)\)
\(=-\left(1-\dfrac{1}{2011}\right)\)
\(=-\dfrac{2010}{2011}\)
