Ta có: \(\frac{1}{1\cdot99}+\frac{1}{3\cdot97}+\cdots+\frac{1}{97\cdot3}+\frac{1}{99\cdot1}\)

\(=2\left(\frac{1}{1\cdot99}+\frac{1}{3\cdot97}+\cdots+\frac{1}{49\cdot51}\right)\)

\(=\frac{2}{100}\left(\frac{100}{1\cdot99}+\frac{100}{3\cdot97}+\cdots+\frac{100}{49\cdot51}\right)\)

\(=\frac{2}{100}\left(1+\frac{1}{99}+\frac13+\frac17+\cdots+\frac{1}{49}+\frac{1}{51}\right)\)

\(=\frac{1}{50}\left(1+\frac13+\cdots+\frac{1}{99}\right)\)

Ta có: \(E=\frac{\left(1+\frac13+\frac15+\cdots+\frac{1}{99}\right)}{\frac{1}{1\cdot99}+\frac{1}{3\cdot97}+\cdots+\frac{1}{99\cdot1}}\)

\(=\frac{\left(1+\frac13+\frac15+\cdots+\frac{1}{99}\right)}{\frac{1}{50}\left(1+\frac13+\frac15+\cdots+\frac{1}{99}\right)}=1:\frac{1}{50}=50\)

\(M=\frac{1}{99.97}-\frac{1}{97.95}-\frac{1}{95.93}-...-\frac{1}{5.3}-\frac{1}{3.1}.\)

\(M=-\left(\frac{1}{1.3}+\frac{1}{3.5}+...+\frac{1}{93.95}+\frac{1}{95.97}+\frac{1}{97.99}\right)\)

\(M=-\frac{1}{2}.\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+...+\frac{1}{93}-\frac{1}{95}+\frac{1}{95}-\frac{1}{97}+\frac{1}{97}-\frac{1}{99}\right)\)

\(M=-\frac{1}{2}.\left(1-\frac{1}{99}\right)=-\frac{1}{2}\cdot\frac{98}{99}=-\frac{49}{99}\)

-98/99

Câu hỏi của Phương Thảo - Toán lớp 7 | Học trực tuyến