`\color{red}{@\text{Mon}}`
`A=1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+\frac{1}{5}-\frac{1}{6}+...+\frac{1}{49}-\frac{1}{50}`
`=1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{45}+\frac{1}{50}-2(\frac{1}{2}+\frac{1}{4}+\frac{1}{6}+...+\frac{1}{50})`
`=\frac{1}{26}+\frac{1}{27}+\frac{1}{28}+...+\frac{1}{50}`
`=(\frac{1}{26}+\frac{1}{27}+\frac{1}{28}+\frac{1}{29}+\frac{1}{30})+(\frac{1}{31}+...+\frac{1}{40})+(\frac{1}{41}+...+\frac{1}{50})`
`\text{ Ta thấy:}` `\frac{1}{26}+\frac{1}{27}+\frac{1}{28}+\frac{1}{29}+\frac{1}{30}>\frac{1}{30}+\frac{1}{30}+\frac{1}{30}+\frac{1}{30}+\frac{1}{30}`
`\text{ Tương tự:}` `\frac{1}{31}+...+\frac{1}{40}>\frac{1}{40}+\frac{1}{40}+...\frac{1}{40}`
`\frac{1}{41}+...+\frac{1}{50}>\frac{1}{50}+\frac{1}{50}+...+\frac{1}{50}`
`\text{ Suy ra:}` `A>5.\frac{1}{30}+10.\frac{1}{40}+10.\frac{1}{50}>\frac{1}{6}+\frac{1}{4}+\frac{1}{5}=\frac{37}{60}`
`Mà` `\frac{37}{30}>\frac{35}{60}=\frac{7}{12}`
`\text{ Theo tính chất bắc cầu:}` `A>\frac{7}{12}(1)`
`\text{ Mật khác ta có:}`
`\frac{1}{26}+\frac{1}{27}+\frac{1}{28}+\frac{1}{29}+\frac{1}{30}<\frac{1}{25}+\frac{1}{25}+\frac{1}{25}+\frac{1}{25}=\frac{1}{25}`
`\text{ Tương tự:}` `\frac{1}{31}+...+\frac{1}{40}<\frac{1}{30}+...+\frac{1}{30}`
`\frac{1}{41}+...+\frac{1}{50}<\frac{1}{40}+...+\frac{1}{40}`
`=>A<5.\frac{1}{25}+10.\frac{1}{30}+10.\frac{1}{40}<\frac{1}{5}+\frac{1}{2}+\frac{1}{4}`
`=\frac{47}{60}<\frac{50}{60}=\frac{5}{6}`
`=>A<\frac{5}{6}(2)`
`\text{ Kết hợp (1) và (2):}`
`=>\frac{7}{12}<A<\frac{5}{6}`
`=>đpcm`
