Question

[Cô Bé Thiên Bình Thùy Linh]

Ngày 31-3-2018

A=\(\frac{4}{2.5}\) +\(\frac{4}{5.8}\) +\(\frac{4}{8.11}\) +.................+\(\frac{4}{65.68}\)

Nhanh tay lên mọi người ơi

Answer 1

[Hoàng Ninh]

\(A=\frac{4}{2.5}+\frac{4}{5.8}+\frac{4}{8.11}+........+\frac{4}{65.68}\)

\(A=4\left(\frac{1}{2.5}+\frac{1}{5.8}+\frac{1}{8.11}+......+\frac{1}{65.68}\right)\)

\(A=\frac{4}{3}\left(\frac{3}{2.5}+\frac{3}{5.8}+\frac{3}{8.11}+..........+\frac{3}{65.68}\right)\)

\(A=\frac{4}{3}\left(\frac{1}{2}-\frac{1}{5}+\frac{1}{5}-\frac{1}{8}+\frac{1}{8}-\frac{1}{11}+\frac{1}{11}-.........-\frac{1}{68}\right)\)

\(A=\frac{4}{3}\left(\frac{1}{2}-\frac{1}{68}\right)\)

\(A=\frac{4}{3}\left(\frac{34}{68}-\frac{1}{68}\right)\)

\(A=\frac{4}{3}.\frac{33}{68}\)

\(A=\frac{11}{17}\)