Đặt \(A=\left(1+\frac{1}{3}+\frac{1}{5}+...+\frac{1}{99}\right):\left(\frac{1}{1.99}+\frac{1}{3.97}+....+\frac{1}{97.3}+\frac{1}{99.1}\right)\)
Đặt \(B=\frac{1}{1.99}+\frac{1}{3.97}+...+\frac{1}{97.3}+\frac{1}{99.1}\)
=> 100 x B = \(\frac{100}{1.99}+\frac{100}{3.97}+...+\frac{100}{97.3}+\frac{100}{99.1}=1+\frac{1}{99}+\frac{1}{3}+\frac{1}{97}+...+\frac{1}{97}+\frac{1}{3}+\frac{1}{99}+1\)
=> 100 x B = \(2.\left(1+\frac{1}{3}+\frac{1}{5}+...+\frac{1}{97}+\frac{1}{99}\right)\)
=> \(B=\frac{1}{50}.\left(1+\frac{1}{3}+\frac{1}{5}+...+\frac{1}{97}+\frac{1}{99}\right)\)
Khi đó A = \(\frac{1+\frac{1}{3}+\frac{1}{5}+...+\frac{1}{97}+\frac{1}{99}}{\frac{1}{50}\left(1+\frac{1}{3}+\frac{1}{5}+...+\frac{1}{97}+\frac{1}{99}\right)}=\frac{1}{\frac{1}{50}}=50\)
