\(\frac{4}{7}x+\left(\frac{1}{1.2.3}+\frac{1}{2.3.4}+...+\frac{1}{12.13.14}\right)=\frac{39}{40}\)
\(\Leftrightarrow\frac{4}{7}x+\left[\frac{1}{2}.\left(\frac{1}{1.2}-\frac{1}{2.3}+\frac{1}{2.3}-\frac{1}{3.4}+...+\frac{1}{12.13}-\frac{1}{13.14}\right)\right]=\frac{39}{40}\)
\(\Leftrightarrow\frac{4}{7}x+\left[\frac{1}{2}.\left(\frac{1}{1.2}-\frac{1}{13.14}\right)\right]=\frac{39}{40}\)
\(\Leftrightarrow\frac{4}{7}x+\left[\frac{1}{2}.\frac{45}{91}\right]=\frac{39}{40}\)
\(\Leftrightarrow\frac{4}{7}x+\frac{45}{182}=\frac{39}{40}\)
\(\Leftrightarrow\frac{4}{7}x=\frac{39}{40}-\frac{45}{182}\Leftrightarrow\frac{4}{7}x=\frac{2649}{3640}\)
\(\Rightarrow x=\frac{2649}{3640}\div\frac{4}{7}=\frac{2649}{2080}\)
Vậy x = \(\frac{2649}{2080}\)
