Giải thích các bước giải:

$\begin{aligned}
& \text{Ta có: } P = \dfrac{1}{1\cdot 2} + \dfrac{1}{3\cdot 4} + \dfrac{1}{5\cdot 6} + \dots + \dfrac{1}{99\cdot 100}\\
& P = 1 - \dfrac{1}{2} + \dfrac{1}{3} - \dfrac{1}{4} + \dfrac{1}{5} - \dfrac{1}{6} + \dots + \dfrac{1}{99} - \dfrac{1}{100} \\
& P = \left(1 + \dfrac{1}{3} + \dfrac{1}{5} + \dots + \dfrac{1}{99}\right) - \left(\dfrac{1}{2} + \dfrac{1}{4} + \dfrac{1}{6} + \dots + \dfrac{1}{100}\right) \\
& P = \left(1 + \dfrac{1}{2} + \dfrac{1}{3} + \dfrac{1}{4} + \dots + \dfrac{1}{99} + \dfrac{1}{100}\right) - 2\left(\dfrac{1}{2} + \dfrac{1}{4} + \dfrac{1}{6} + \dots + \dfrac{1}{100}\right) \\
& P = \left(1 + \dfrac{1}{2} + \dfrac{1}{3} + \dots + \dfrac{1}{100}\right) - \left(1 + \dfrac{1}{2} + \dfrac{1}{3} + \dots + \dfrac{1}{50}\right) \\
& P = \dfrac{1}{51} + \dfrac{1}{52} + \dfrac{1}{53} + \dots + \dfrac{1}{100} \\
& P = \left(\dfrac{1}{51} + \dfrac{1}{52} + \dots + \dfrac{1}{75}\right) + \left(\dfrac{1}{76} + \dfrac{1}{77} + \dots + \dfrac{1}{100}\right) \\
& \dfrac{1}{51} + \dfrac{1}{52} + \dots + \dfrac{1}{75} > \dfrac{1}{75} + \dfrac{1}{75} + \dots + \dfrac{1}{75} \\
& \dfrac{1}{51} + \dfrac{1}{52} + \dots + \dfrac{1}{75} > 25 \cdot \dfrac{1}{75} \\
& \dfrac{1}{51} + \dfrac{1}{52} + \dots + \dfrac{1}{75} > \dfrac{1}{3} \\
& \dfrac{1}{76} + \dfrac{1}{77} + \dots + \dfrac{1}{100} > \dfrac{1}{100} + \dfrac{1}{100} + \dots + \dfrac{1}{100} \\
& \dfrac{1}{76} + \dfrac{1}{77} + \dots + \dfrac{1}{100} > 25 \cdot \dfrac{1}{100} \\
& \dfrac{1}{76} + \dfrac{1}{77} + \dots + \dfrac{1}{100} > \dfrac{1}{4} \\
& P > \dfrac{1}{3} + \dfrac{1}{4} \\
& P > \dfrac{7}{12} \\
& \dfrac{1}{51} + \dfrac{1}{52} + \dots + \dfrac{1}{75} < \dfrac{1}{50} + \dfrac{1}{50} + \dots + \dfrac{1}{50} \\
& \dfrac{1}{51} + \dfrac{1}{52} + \dots + \dfrac{1}{75} < 25 \cdot \dfrac{1}{50} \\
& \dfrac{1}{51} + \dfrac{1}{52} + \dots + \dfrac{1}{75} < \dfrac{1}{2} \\
& \dfrac{1}{76} + \dfrac{1}{77} + \dots + \dfrac{1}{100} < \dfrac{1}{75} + \dfrac{1}{75} + \dots + \dfrac{1}{75} \\
& \dfrac{1}{76} + \dfrac{1}{77} + \dots + \dfrac{1}{100} < 25 \cdot \dfrac{1}{75} \\
& \dfrac{1}{76} + \dfrac{1}{77} + \dots + \dfrac{1}{100} < \dfrac{1}{3} \\
& P < \dfrac{1}{2} + \dfrac{1}{3} \\
& P < \dfrac{5}{6} \\
& \text{Kết quả: } \dfrac{7}{12} < P < \dfrac{5}{6}
\end{aligned}$