\(=1-\left(\frac{2}{1.3}-\frac{2}{3.5}-...-\frac{2}{2005-2007}\right)\)
\(=1-\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+...+\frac{1}{2005}-\frac{1}{2007}\right)\)
\(=1-\left[1+\left(-\frac{1}{3}+\frac{1}{3}\right)+\left(-\frac{1}{5}+\frac{1}{5}\right)+...+\left(-\frac{1}{2005}+\frac{1}{2005}\right)-\frac{1}{2007}\right]\)
\(=1-\left(1+0+0+...+0-\frac{1}{2007}\right)\)
\(=1-\left(1-\frac{1}{2007}\right)\)
\(=1-1+\frac{1}{2007}\)
\(=0+\frac{1}{2007}\)
\(=\frac{1}{2007}\)
Ai thấy tớ đúng k nha
Đặt A = \(1-\frac{2}{1.3}-\frac{2}{3.5}-.....-\frac{2}{2005.2007}\)
= \(1-\left(\frac{2}{1.3}+\frac{2}{3.5}+....+\frac{2}{2005.2007}\right)\)
=\(1-\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+....+\frac{1}{2005}-\frac{1}{2007}\right)\)
= \(1-\left(1-\frac{1}{2017}\right)\)
=\(1-1+\frac{1}{2017}\)
=\(0+\frac{1}{2017}\)
=\(\frac{1}{2017}\)
