chỉnh đề

\(A=\frac{4}{1.3}+\frac{4}{3.5}+\frac{4}{5.7}+....+\frac{4}{2015.2017}\)

\(=2\left(\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+....+\frac{2}{2015.2017}\right)\)

\(=2\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+....+\frac{1}{2015}-\frac{1}{2017}\right)\)

\(=2\left(1-\frac{1}{2017}\right)\)

\(=2.\frac{2016}{2017}=\frac{4032}{2017}\)

p/s: chúc bạn học tốt

\(A=\frac{4}{1.3}+\frac{4}{3.5}+...+\frac{4}{2015.2017}\)

\(A=2\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+...+\frac{1}{2015}-\frac{1}{2017}\right)\)

\(A=2\left(1-\frac{1}{2017}\right)\)

\(A=\frac{2016.2}{2017}\)

\(A=\frac{4}{1.3}+\frac{4}{3.5}+\frac{4}{5.7}+...+\frac{4}{2015.2017}\)

\(=2\left(\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+...+\frac{2}{2015.2017}\right)\)

\(=2\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{2015}-\frac{1}{2017}\right)\)

\(=2\left[1+\left(-\frac{1}{3}+\frac{1}{3}\right)+\left(-\frac{1}{5}+\frac{1}{5}\right)+\left(-\frac{1}{7}+\frac{1}{7}\right)+...+\left(-\frac{1}{2015}+\frac{1}{2015}\right)-\frac{1}{2017}\right]\)

\(=2\left(1+0+0+0+...+0-\frac{1}{2017}\right)\)

\(=2\left(1-\frac{1}{2017}\right)=2\left(\frac{2017}{2017}-\frac{1}{2017}\right)\)

\(=2.\frac{2016}{2017}=\frac{4032}{2017}\)