\(A=a.\frac{1}{2} +a.\frac{1}{3}-a.\frac{1}{4}=a.\left ( \frac{1}{2}+\frac{1}{3}-\frac{1}{4} \right )=\frac{7}{12}a\)

Thay \(a= \frac{-4}{5}\) ta được:

\(A=\frac{7}{12}.\frac{-4}{5}=\frac{7.(-4)}{12.5}=\frac{-7}{15}\)

\(B=\frac{3}{4}.b+\frac{4}{3}.b-\frac{1}{2}.b=b.\left ( \frac{3}{4}+\frac{4}{3}-\frac{1}{2} \right )=\frac{19}{12}b\)

Thay \(b=\frac{6}{19}\) ta được

\(B=\frac{19}{12}.\frac{6}{19}=\frac{6}{12}=\frac{1}{2}\)

\(C=c.\frac{3}{4}+c.\frac{5}{6}-c.\frac{19}{12}=c.\left ( \frac{3}{4}+\frac{5}{6}-\frac{19}{12}=0.c \right )\)

Thay \(c=\frac{2002}{2003}\) ta được

$C=0.\frac{2002}{2003}=0$