(1+2+3+...+2009)(12.6-36.2):\(\left(\frac{1}{10}+\frac{1}{11}+\frac{1}{12}\right)\)
=(1+2+3+...+2009)(12.3.2-12.3.2):\(\left(\frac{1}{10}+\frac{1}{11}+\frac{1}{12}\right)\)
=(1+2+3+...+2009).0:\(\left(\frac{1}{10}+\frac{1}{11}+\frac{1}{12}\right)\)
=0