Đặt \(A=\frac{1}{50.48}-\frac{1}{48.46}-...-\frac{1}{4.2}\) ta có :
\(A=\frac{1}{48.50}-\left(\frac{1}{2.4}+\frac{1}{4.6}+\frac{1}{6.8}+...+\frac{1}{46.48}\right)\) ( xắp sếp lại cho đẹp đội hình thôi :)
Đặt \(B=\frac{1}{2.4}+\frac{1}{4.6}+\frac{1}{6.8}+...+\frac{1}{46.48}\) ta có :
\(2B=\frac{2}{2.4}+\frac{2}{4.6}+\frac{2}{6.8}+...+\frac{2}{46.48}\)
\(2B=\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-\frac{1}{6}+\frac{1}{6}-\frac{1}{8}+...+\frac{1}{46}-\frac{1}{48}\)
\(2B=\frac{1}{2}-\frac{1}{48}\)
\(2B=\frac{23}{48}\)
\(B=\frac{23}{48}:2\)
\(B=\frac{23}{48}.\frac{1}{2}\)
\(B=\frac{23}{96}\)
\(\Rightarrow\)\(A=\frac{1}{48.50}-B=\frac{1}{48.50}-\frac{23}{96}=\frac{1}{2400}-\frac{23}{96}=\frac{-287}{1200}\)
Vậy \(A=\frac{-287}{1200}\)
Chúc bạn học tốt ~
