Ta có: \(C=c\cdot\dfrac{3}{4}+c\cdot\dfrac{5}{6}-c\cdot\dfrac{19}{12}\)
\(=c\cdot\left(\dfrac{3}{4}+\dfrac{5}{6}-\dfrac{19}{12}\right)\)
\(=c\cdot\left(\dfrac{9}{12}+\dfrac{10}{12}-\dfrac{19}{12}\right)\)
\(=0\)
Vậy: Khi \(c=\dfrac{2002}{2003}\) thì C=0