Ta có:
\(C=\frac{1.5.6+2.10.12+4.20.24+...+9.45.54}{1.3.5+2.6.10+4.12.20+...+9.27.45}\)
\(=\frac{1.3.5.2+2.6.10.2+4.12.20.2+...+9.27.45.2}{1.3.5+2.6.10+4.12.20+...+9.27.45}\)
\(=\frac{\left(1.3.5+2.6.10+4.12.20+...+9.27.45\right).2}{1.3.5+2.6.10+4.12.20+...+9.27.45}\)
\(=2\)
Vậy C=2.