\(A=\frac{17}{3}\left(\frac{6}{7.13}+\frac{9}{13.22}+\frac{15}{22.37}+\frac{12}{37.49}\right)=\frac{17}{3}\left(\frac{1}{7}-\frac{1}{13}+......-\frac{1}{49}\right)=\frac{17}{3}\left(\frac{1}{7}-\frac{1}{49}\right);B=\frac{13}{3}\left(\frac{9}{7.16}+\frac{15}{16.31}+\frac{12}{31.43}+\frac{6}{43.49}\right)=\frac{13}{3}\left(\frac{1}{7}-\frac{1}{16}+\frac{1}{16}-.....-\frac{1}{49}\right)=\frac{13}{3}\left(\frac{1}{7}-\frac{1}{49}\right)\Rightarrow\frac{A}{B}=\frac{17}{13}\)
Có: \(A=\frac{17}{3}\left(\frac{6}{7.13}+\frac{9}{13.22}+\frac{15}{22.37}+\frac{12}{37.49}\right)\)
\(A=\frac{17}{3}\left(\frac{1}{7}-\frac{1}{13}+\frac{1}{13}-\frac{1}{22}+\frac{1}{22}-\frac{1}{37}+\frac{1}{37}-\frac{1}{49}\right)\)
\(A=\frac{17}{3}\left(\frac{1}{7}-\frac{1}{49}\right)\)
Ttự, ta đc: \(B=\frac{13}{3}\left(\frac{1}{7}-\frac{1}{49}\right)\)
Vậy \(\frac{A}{B}=\frac{\frac{17}{3}}{\frac{13}{3}}=\frac{17}{13}\)
#Walker