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Câu hỏi của Nguyễn Thị Ngọc Ánh - Toán lớp 7 - Học toán với OnlineMath

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Ta có : 

\(A=\frac{1}{2003\cdot2002}-\frac{1}{2002\cdot2001}-...-\frac{1}{3\cdot2}-\frac{1}{2\cdot1}\)

\(A=-\left(\frac{1}{2003\cdot2002}+\frac{1}{2002\cdot2001}+...+\frac{1}{3\cdot2}+\frac{1}{2\cdot1}\right)\)

\(A=-\left(\frac{1}{1\cdot2}+\frac{1}{2\cdot3}+...+\frac{1}{2001\cdot2002}+\frac{1}{2002\cdot2003}\right)\)

\(A=-\left(\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{2001}-\frac{1}{2002}+\frac{1}{2002}-\frac{1}{2003}\right)\)

\(A=-\left(1-\frac{1}{2003}\right)\)

\(A=-\frac{2002}{2003}\)

\(A=\frac{1}{2003.2002}-\frac{1}{2002.2001}-...-\frac{1}{3.2}-\frac{1}{2.1}\)

\(=-\left(\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{2001.2002}\right)+\frac{1}{2002}.\frac{1}{2003}\)

\(=-\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{2001}-\frac{1}{2002}\right)+\frac{1}{2002}.\frac{1}{2003}\)

\(=-\left(1-\frac{1}{2002}\right)+\frac{1}{2002}.\frac{1}{2003}\)

\(=-1+\frac{1}{2002}.+\frac{1}{2002}.\frac{1}{2003}\)

\(=-1+\frac{1}{2002}\left(1+\frac{1}{2003}\right)\)

\(=-1+\frac{1}{2002}.\frac{2004}{2003}\)

\(=-1+\frac{2}{2003}\)

\(=\frac{-2003+2}{2003}\)

\(=\frac{-2001}{2003}\)

\(-\frac{1}{2003\cdot2002}-\frac{1}{2002\cdot2001}-\frac{1}{2001\cdot2000}-...-\frac{1}{2\cdot1}\) 

\(=-1\left(\frac{1}{1\cdot2}+...+\frac{1}{2000\cdot2001}+\frac{1}{2001\cdot2002}+\frac{1}{2002\cdot2003}\right)\) 

\(=-1\left(\frac{1}{1}-\frac{1}{2}+...+\frac{1}{2000}-\frac{1}{2001}+\frac{1}{2001}-\frac{1}{2002}+\frac{1}{2002}-\frac{1}{2003}\right)\) 

\(=-1\left(1-\frac{1}{2003}\right)\) 

\(=-1\left(\frac{2003}{2003}-\frac{1}{2003}\right)\)              

\(=-1\cdot\frac{2002}{2003}\) 

\(=-\frac{2002}{2003}\)

\(\frac{1}{2003.2002}-\frac{1}{2002.2001}-...-\frac{1}{2.1}\)

\(=\frac{1}{2003.2002}-\left(\frac{1}{1.2}+\frac{1}{3.2}+...+\frac{1}{2001.2002}\right)\)

\(=\frac{1}{2003.2002}-\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-...+\frac{1}{2000}-\frac{1}{2001}+\frac{1}{2001}-\frac{1}{2002}\right)\)

\(=\frac{1}{2003.2002}-\left(1-\frac{1}{2002}\right)\)

\(=\frac{1}{2003.2002}-\frac{2001}{2002}\)

\(\frac{1}{2003.2002}-\frac{1}{2002.2001}-...-\frac{1}{2.1}\)

\(=\frac{1}{2003.2002}-\left(\frac{1}{2002.2001}+\frac{1}{2001.2000}+...+\frac{1}{2.1}\right)\)

\(=\frac{1}{2003.2002}-\left(\frac{1}{2002}-\frac{1}{2001}+\frac{1}{2001}-\frac{1}{2000}+...+\frac{1}{2}-1\right)\)

\(=\frac{1}{2003.2002}-\left(\frac{1}{2002}-1\right)\)

\(=\frac{1}{2003.2002}-\left(\frac{1}{2002}-\frac{2002}{2002}\right)\)

\(=\frac{1}{2003.2002}-\frac{-2001}{2002}\)

\(=\frac{1}{2003}-\frac{1}{2002}+\frac{2001}{2002}\)

\(=\frac{1}{2003}+\frac{2000}{2002}\)

\(=\frac{1}{2003}+\frac{1000}{1001}\)

Ko chắc ạ! Bạn nào rãnh thì check hoặc nhận xét hộ!

\(E=\frac{2}{3.5}+\frac{7}{5.12}+\frac{9}{4.39}=\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{12}+\frac{27}{12.39}=\frac{1}{3}-\frac{1}{12}+\frac{1}{12}-\frac{1}{39}=\frac{1}{3}-\frac{1}{39}=\frac{4}{13}\)

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