Question

 help

Answer 1

b) \(2^2>2.3\Rightarrow\dfrac{1}{2^2}< \dfrac{1}{2.3}\)\(;3^2>2.3\Rightarrow\dfrac{1}{3^2}< \dfrac{1}{2.3}\);...\(;9^2>8.9\Rightarrow\dfrac{1}{9^2}< \dfrac{1}{8.9}\)

\(\Rightarrow B< \dfrac{1}{1.2}+\dfrac{1}{2.3}+...+\dfrac{1}{8.9}\)\(\Rightarrow B< 1-\dfrac{1}{2}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{8}-\dfrac{1}{9}\)

\(\Rightarrow B< 1-\dfrac{1}{9}=\dfrac{8}{9}\) (1)

Mặt khác, ta có \(2^2< 2.3\Rightarrow\dfrac{1}{2^2}>\dfrac{1}{2.3}\);...;\(9^2< 9.10\Rightarrow\dfrac{1}{9^2}>\dfrac{1}{9.10}\)

\(\Rightarrow B>\dfrac{1}{2.3}+\dfrac{1}{3.4}+...+\dfrac{1}{9.10}\)\(\Rightarrow B>\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{9}-\dfrac{1}{10}\)

\(\Rightarrow B>\dfrac{1}{2}-\dfrac{1}{10}=\dfrac{2}{5}\) (2)

Từ (1) và (2) \(\Rightarrow\dfrac{2}{5}< B< \dfrac{8}{9}\)