Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
Ta thấy:
\(\dfrac{1}{51}< \dfrac{1}{50}\)
\(\dfrac{1}{52}< \dfrac{1}{50}\)
...
\(\dfrac{1}{100}< \dfrac{1}{50}\)
\(\Rightarrow\dfrac{1}{51}+\dfrac{1}{52}+...+\dfrac{1}{100}< \dfrac{1}{50}.50=1\)
\(\Rightarrow\dfrac{1}{51}+\dfrac{1}{52}+...+\dfrac{1}{100}< 1\left(1\right)\)
Lại có:
\(\dfrac{1}{51}>\dfrac{1}{100}\)
\(\dfrac{1}{52}>\dfrac{1}{100}\)
...
\(\dfrac{1}{100}=\dfrac{1}{100}\)
\(\Rightarrow\dfrac{1}{51}+\dfrac{1}{52}+...+\dfrac{1}{100}>\dfrac{1}{100}.50=\dfrac{1}{2}\)
\(\Rightarrow\dfrac{1}{51}+\dfrac{1}{52}+...+\dfrac{1}{100}>\dfrac{1}{2}\left(2\right)\)
Từ (1),(2)\(\Rightarrow\)\(\dfrac{1}{2}< \dfrac{1}{51}+\dfrac{1}{52}+...+\dfrac{1}{100}< 1\)
a) $A=\dfrac{1}{1.2}+\dfrac{1}{3.4}+...+\dfrac{1}{99.100}$
$=>A=1-\dfrac{1}{2}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{99}-\dfrac{1}{100}$
$=>A=(1+\dfrac{1}{3}+...+\dfrac{1}{99})-(\dfrac{1}{2}+\dfrac{1}{4}+...+\dfrac{1}{100})$
$=>A=(1+\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+...+\dfrac{1}{99}+\dfrac{1}{100})-(\dfrac{1}{2}+\dfrac{1}{4}+...+\dfrac{1}{100}.2)$
$=>A=(1+\dfrac{1}{2}+\dfrac{1}{3}+...+\dfrac{1}{100})-(1+\dfrac{1}{2}+...+\dfrac{1}{50})$
$=>A=\dfrac{1}{51}+\dfrac{1}{52}+...+\dfrac{1}{100}$
b) Ta có : $A=1-\dfrac{1}{2}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{99}-\dfrac{1}{100}$
$=>A=(1-\dfrac{1}{2}+\dfrac{1}{3})-(\dfrac{1}{4}-\dfrac{1}{5})-...-(\dfrac{1}{98}-\dfrac{1}{99})-\dfrac{1}{100}$
$=>A<1-\dfrac{1}{2}+\dfrac{1}{3}=\dfrac{5}{6}$
a=\(\dfrac{1}{2}\cdot\dfrac{3}{4}\cdot\dfrac{5}{6}\cdot\dfrac{7}{8}\cdot...\cdot\dfrac{79}{80}\)
a<\(\dfrac{2}{3}\cdot\dfrac{4}{5}\cdot\dfrac{6}{7}\cdot\dfrac{8}{9}\cdot...\cdot\dfrac{80}{81}\)
\(\text{a}^2< \dfrac{1}{2}\cdot\dfrac{2}{3}\cdot\dfrac{3}{4}\cdot\dfrac{4}{5}\cdot\dfrac{5}{6}\cdot\dfrac{6}{7}\cdot\dfrac{7}{8}\cdot\dfrac{8}{9}\cdot...\cdot\dfrac{79}{80}\cdot\dfrac{80}{81}\)
\(\Rightarrow\text{a}^2< \dfrac{1}{81}=\left(\dfrac{1}{9}\right)^2\)
\(\Rightarrow\text{a}< \dfrac{1}{9}\)(dpcm)
Nho tich cho mk nhe
Đặt \(A=\dfrac{1}{51}+\dfrac{1}{52}+...+\dfrac{1}{100}\)
Ta có: \(\dfrac{1}{51}+\dfrac{1}{52}+...+\dfrac{1}{75}>\dfrac{1}{75}+\dfrac{1}{75}+...+\dfrac{1}{75}=\dfrac{25}{75}=\dfrac{1}{3}\)
\(\dfrac{1}{76}+\dfrac{1}{77}+...+\dfrac{1}{100}>\dfrac{1}{100}+\dfrac{1}{100}+...+\dfrac{1}{100}=\dfrac{25}{100}=\dfrac{1}{4}\)
Do đó: \(A>\dfrac{1}{3}+\dfrac{1}{4}=\dfrac{7}{12}\)(1)
Ta có: \(\dfrac{1}{51}+\dfrac{1}{52}+...+\dfrac{1}{75}< \dfrac{1}{50}+\dfrac{1}{50}+...+\dfrac{1}{50}=\dfrac{25}{50}=\dfrac{1}{2}\)
\(\dfrac{1}{76}+\dfrac{1}{77}+...+\dfrac{1}{100}< \dfrac{1}{75}+\dfrac{1}{75}+...+\dfrac{1}{75}=\dfrac{25}{75}=\dfrac{1}{3}\)
Do đó: \(A< \dfrac{1}{2}+\dfrac{1}{3}=\dfrac{5}{6}\)(2)
Từ (1) và (2) ta suy ra ĐPCM