\(A=\dfrac{3}{3.5} + \dfrac{3}{5.7} + ... + \dfrac{3}{97.99}\)
\(\Rightarrow A=\dfrac{3}{2}\left(\dfrac{2}{3.5}+\dfrac{2}{5.7}+...+\dfrac{2}{97.99}\right)\)
\(\Rightarrow A=\dfrac{3}{2}\left(\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+...+\dfrac{1}{97}-\dfrac{1}{99}\right)\)
\(\Rightarrow A=\dfrac{3}{2}\left(\dfrac{1}{3}-\dfrac{1}{99}\right)\)
\(\Rightarrow A=\dfrac{3}{2}.\dfrac{32}{99}\)
\(\Rightarrow A=\dfrac{16}{33}\)
Vậy \(A=\dfrac{16}{33}\)
A= \(\dfrac{3}{3.5}+\dfrac{3}{5.7}+\dfrac{3}{7.9}+...+\dfrac{3}{97.99}\)
= \(\dfrac{3}{2}.\left(\dfrac{1}{3.5}+\dfrac{1}{5.7}+\dfrac{1}{7.9}+...+\dfrac{1}{97.99}\right)\)
= \(\dfrac{3}{2}.\left(\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{9}+...+\dfrac{1}{97}-\dfrac{1}{99}\right)\)
= \(\dfrac{3}{2}.\left(\dfrac{1}{3}-\dfrac{1}{99}\right)\)
= \(\dfrac{3}{2}.\dfrac{32}{99}\)
= \(\dfrac{3.32}{2.99}\)= \(\dfrac{3.2.3.6}{2.11.3.3}\)= \(\dfrac{6}{11}\)
cho mình hỏi là tại sao lại đặt \(\dfrac{3}{2}\) ra ngoài ?
