Sửa đề
\(B=\frac{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{2017}}{\frac{2016}{1}+\frac{1015}{2}+\frac{2014}{3}+...+\frac{1}{2016}}\)
\(\Leftrightarrow B=\frac{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+....+\frac{1}{2017}}{\left(\frac{2015}{2}+1\right)+\left(\frac{2014}{3}+1\right)+...+\left(\frac{1}{2016}+1\right)+1}\)
\(\Leftrightarrow B=\frac{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{2017}}{\frac{2017}{2}+\frac{2017}{3}+...+\frac{2017}{2016}+\frac{2017}{2017}}\)
\(\Leftrightarrow B=\frac{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{2017}}{2017.\left(\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2016}+\frac{1}{2017}\right)}\)
\(\Leftrightarrow\) \(B=\frac{1}{2017}\)
Vậy \(B=\frac{1}{2017}\)
@@ Học tốt @@
Chiyuki Fujito
