c) \(\frac{x-\frac{1}{2}}{\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+\frac{1}{20}}=\frac{1}{3}\)
Đặt \(S=\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+\frac{1}{20}\)
\(S=\frac{1}{1\cdot2}+\frac{1}{2\cdot3}+\frac{1}{3\cdot4}+\frac{1}{4\cdot5}\)
\(S=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{4}-\frac{1}{5}\)
\(S=1-\frac{1}{5}=\frac{4}{5}\)
Do đó \(\frac{x-\frac{1}{2}}{S}=\frac{1}{3}\)
=> \(x-\frac{1}{2}=\frac{1}{3}S=\frac{1}{3}\cdot\frac{4}{5}=\frac{4}{15}\)
=> \(x=\frac{4}{15}+\frac{1}{2}=\frac{23}{30}\)