\(\frac{1}{5}A=\frac{4}{31.35}+\frac{6}{35.41}+\frac{9}{41.50}+\frac{7}{50.57}\)
\(A=\frac{1}{31}-\frac{1}{35}+\frac{1}{35}-\frac{1}{41}+\frac{1}{41}-\frac{1}{50}+\frac{1}{50}-\frac{1}{57}=\frac{1}{31}-\frac{1}{57}\)
\(\frac{1}{2}B=\frac{7}{31.38}+\frac{5}{38.43}+\frac{3}{43.46}+\frac{11}{46.57}=\frac{1}{31}-\frac{1}{38}+\frac{1}{38}-\frac{1}{43}+...+\frac{1}{46}-\frac{1}{57}=\frac{1}{31}-\frac{1}{57}\)
Vậy \(\frac{1}{5}A=\frac{1}{2}B\) hay \(\frac{A}{B}=\frac{1}{2}:\frac{1}{5}=\frac{5}{2}\)