Question

Tính nhanh: \(\dfrac{2}{3}+\dfrac{2}{6}+\dfrac{2}{12}+\dfrac{2}{24}+\dfrac{2}{48}+\dfrac{2}{96}+\dfrac{2}{192}\)

Answer 1

Đặt \(A=\dfrac{2}{3}+\dfrac{2}{6}+\dfrac{2}{12}+\dfrac{2}{24}+\dfrac{2}{48}+\dfrac{2}{98}+\dfrac{2}{192}\)

\(2A=\dfrac{4}{3}+\dfrac{2}{3}+\dfrac{2}{6}+\dfrac{2}{12}+\dfrac{2}{24}+\dfrac{2}{98}+\dfrac{2}{192}\)

\(2A-A=\left(\dfrac{4}{3}+\dfrac{2}{3}+\dfrac{2}{6}+\dfrac{2}{12}+\dfrac{2}{24}+\dfrac{2}{98}+\dfrac{2}{192}\right)-\left(\dfrac{2}{3}+\dfrac{2}{6}+\dfrac{2}{12}+\dfrac{2}{24}+\dfrac{2}{48}+\dfrac{2}{98}+\dfrac{2}{192}\right)\)

\(A=\dfrac{4}{3}-\dfrac{2}{192}\)

\(A=\dfrac{127}{96}\)

Vậy \(A=\dfrac{127}{96}\)