\(x+\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+\frac{1}{30}=\frac{47}{42}\)
\(x+\left(\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+\frac{1}{30}\right)=\frac{47}{42}\)
\(x+A=\frac{47}{42}\)
ta thấy :
\(A=\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+\frac{1}{30}\)
\(A=\frac{1}{1\cdot2}+\frac{1}{2\cdot3}+\frac{1}{3\cdot4}+\frac{1}{4\cdot5}+\frac{1}{5\cdot6}\)
\(A=\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}\)
\(A=\frac{1}{1}-\frac{1}{6}\)
\(A=\frac{5}{6}\)
vậy \(x+\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+\frac{1}{30}=\frac{47}{42}\)
hay \(x+\frac{5}{6}=\frac{47}{42}\)
\(x=\frac{47}{42}-\frac{5}{6}\)
\(x=\frac{2}{7}\)
\(x+\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+\frac{1}{30}=\frac{47}{42}\)
\(x=\frac{47}{42}-\left(\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+\frac{1}{30}\right)\)
\(x=\frac{47}{42}-\frac{5}{6}\)
\(x=\frac{2}{7}.\)