Đáp án:
Giải thích các bước giải:
`A=\frac{1}{1.2}+\frac{1}{3.4}+...+\frac{1}{99.100}`
`⇒A=1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+....+\frac{1}{99}-\frac{1}{100}`
`⇒A=1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+....+\frac{1}{99}+\frac{1}{100}-2(\frac{1}{2}+\frac{1}{4}+...+\frac{1}{100})`
`⇒A=1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+....+\frac{1}{99}+\frac{1}{100}-(1+\frac{1}{2}+...+\frac{1}{50})`
`⇒A=\frac{1}{51}+\frac{1}{52}+...+\frac{1}{100}`
`⇒A=(\frac{1}{51}+\frac{1}{52}+...+\frac{1}{75})+(\frac{1}{76}+\frac{1}{77}+...+\frac{1}{100})`
Có `\frac{1}{51}+\frac{1}{52}+...+\frac{1}{75}>\frac{1}{75}.25=\frac{1}{3}`
`\frac{1}{76}+\frac{1}{77}+...+\frac{1}{100}>\frac{1}{100}.25=\frac{1}{4}`
`\frac{1}{51}+\frac{1}{52}+...+\frac{1}{75}<\frac{1}{50}.25=\frac{1}{2}`
`\frac{1}{76}+\frac{1}{77}+...+\frac{1}{100}>\frac{1}{75}.25=\frac{1}{3}`
Nên `\frac{1}{3}+\frac{1}{4}<A<\frac{1}{2}+\frac{1}{3}`
hay `\frac{7}{12}<A<\frac{5}{6}`(đpcm)
