\(A=\frac{45}{7.16}+\frac{75}{16.31}+\frac{60}{31.43}+\frac{135}{43.70}\)

\(=5\left(\frac{1}{7}-\frac{1}{16}+\frac{1}{16}-\frac{1}{31}+\frac{1}{31}-\frac{1}{43}+\frac{1}{43}-\frac{1}{70}\right)\)

\(=5\left(\frac{1}{7}-\frac{1}{70}\right)\)

mk sửa đề B

\(B=\frac{18}{7.13}+\frac{36}{13.25}+\frac{72}{25.49}+\frac{63}{49.70}\)

\(=3\left(\frac{1}{7}-\frac{1}{13}+\frac{1}{13}-\frac{1}{25}+\frac{1}{25}-\frac{1}{49}+\frac{1}{49}-\frac{1}{70}\right)\)

\(=3\left(\frac{1}{7}-\frac{1}{70}\right)\)

Vậy  \(\frac{A}{B}=\frac{5\left(\frac{1}{7}-\frac{1}{70}\right)}{3\left(\frac{1}{7}-\frac{1}{70}\right)}=\frac{5}{3}\)

\(A=\frac{34}{7.13}+\frac{51}{13.22}+\frac{85}{22.37}+\frac{68}{37.49}\)

\(=17.\left(\frac{2}{7.13}+\frac{3}{13.22}+\frac{5}{22.37}+\frac{4}{37.49}\right)\)

\(=\frac{17}{3}.\left(\frac{6}{7.13}+\frac{9}{13.22}+\frac{15}{22.37}+\frac{12}{37.49}\right)\)

\(=\frac{17}{3}\left(\frac{1}{7}-\frac{1}{13}+\frac{1}{13}-\frac{1}{22}+\frac{1}{22}-\frac{1}{37}+\frac{1}{37}-\frac{1}{49}\right)\)

\(=\frac{17}{3}\left(\frac{1}{7}-\frac{1}{49}\right)\)

\(B=\frac{39}{7.16}+\frac{65}{16.31}+\frac{52}{31.43}+\frac{26}{43.49}\)

\(=13\left(\frac{3}{7.16}+\frac{5}{16.31}+\frac{4}{31.43}+\frac{2}{43.49}\right)\)

\(=\frac{13}{3}\left(\frac{9}{7.16}+\frac{15}{16.31}+\frac{12}{31.43}+\frac{6}{43.49}\right)\)

\(=\frac{13}{3}\left(\frac{1}{7}-\frac{1}{16}+\frac{1}{16}-\frac{1}{31}+\frac{1}{31}-\frac{1}{43}+\frac{1}{43}-\frac{1}{49}\right)\)

\(=\frac{13}{3}\left(\frac{1}{7}-\frac{1}{49}\right)\)

\(\Rightarrow\frac{A}{B}=\frac{\frac{17}{3}\left(\frac{1}{7}-\frac{1}{49}\right)}{\frac{13}{3}\left(\frac{1}{7}-\frac{1}{49}\right)}=\frac{\frac{17}{3}}{\frac{13}{3}}=\frac{17}{13}\)

\(A=\frac{34}{7.13}+\frac{51}{13.22}+\frac{85}{22.37}+\frac{68}{37.49}\)

\(A=17.\left(\frac{2}{7.13}+\frac{3}{13.22}+\frac{5}{22.37}+\frac{4}{37.49}\right)\)

\(A=\frac{17}{3}.\left(\frac{6}{7.13}+\frac{9}{13.22}+\frac{15}{22.37}+\frac{12}{37.49}\right)\)

\(A=\frac{17}{3}.\left(\frac{1}{7}-\frac{1}{13}+\frac{1}{13}-\frac{1}{22}+\frac{1}{22}-\frac{1}{37}+\frac{1}{37}-\frac{1}{49}\right)\)

\(A=\frac{17}{3}.\left(\frac{1}{7}-\frac{1}{49}\right)\)

\(A=\frac{17}{3}.\frac{6}{49}\)

\(B=\frac{39}{7.16}+\frac{65}{16.31}+\frac{52}{31.43}+\frac{26}{43.49}\)

\(B=13.\left(\frac{3}{7.16}+\frac{5}{16.31}+\frac{4}{31.43}+\frac{2}{43.49}\right)\)

\(B=\frac{13}{3}.\left(\frac{9}{7.16}+\frac{15}{16.31}+\frac{12}{31.43}+\frac{4}{43.49}\right)\)

\(B=\frac{13}{3}.\left(\frac{1}{7}-\frac{1}{16}+\frac{1}{16}-\frac{1}{31}+\frac{1}{31}-\frac{1}{43}+\frac{1}{43}-\frac{1}{49}\right)\)

\(B=\frac{13}{3}.\left(\frac{1}{7}-\frac{1}{49}\right)=\frac{13}{3}.\frac{6}{49}\)

\(\frac{A}{B}=\frac{\frac{17}{3}.\frac{6}{49}}{\frac{13}{3}.\frac{6}{49}}=\frac{17}{13}\)

đoạn cuối bạn kia làm sai một chỗ

tính 3A rồi => A = 34/49 

tính 3B => B = 26/49 

=> A/B = 17/13

\(A=\frac{17}{3}\left(\frac{6}{7.13}+\frac{9}{13.22}+\frac{15}{22.37}+\frac{12}{37.49}\right)=\frac{17}{3}\left(\frac{1}{7}-\frac{1}{13}+......-\frac{1}{49}\right)=\frac{17}{3}\left(\frac{1}{7}-\frac{1}{49}\right);B=\frac{13}{3}\left(\frac{9}{7.16}+\frac{15}{16.31}+\frac{12}{31.43}+\frac{6}{43.49}\right)=\frac{13}{3}\left(\frac{1}{7}-\frac{1}{16}+\frac{1}{16}-.....-\frac{1}{49}\right)=\frac{13}{3}\left(\frac{1}{7}-\frac{1}{49}\right)\Rightarrow\frac{A}{B}=\frac{17}{13}\)

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

\(A=17\left(\frac{2}{7\cdot13}+\frac{3}{13\cdot22}+\frac{5}{22\cdot37}+\frac{4}{37\cdot49}\right)\)

\(=\frac{17}{3}\left(\frac{6}{7\cdot13}+\frac{9}{13\cdot22}+\frac{15}{22\cdot37}+\frac{12}{37\cdot49}\right)\)

\(=\frac{17}{3}\left(\frac{1}{7}-\frac{1}{13}+\frac{1}{13}-\frac{1}{22}+\frac{1}{22}-\frac{1}{37}+\frac{1}{37}-\frac{1}{49}\right)\)

\(=\frac{17}{3}\left(\frac{1}{7}-\frac{1}{49}\right)\)

\(B=13\left(\frac{3}{7\cdot16}+\frac{5}{16\cdot31}+\frac{4}{31\cdot43}+\frac{2}{43\cdot49}\right)\)

\(=\frac{13}{3}\left(\frac{9}{7\cdot16}+\frac{15}{16\cdot31}+\frac{12}{31\cdot43}+\frac{6}{43\cdot49}\right)\)  

\(=\frac{13}{3}\left(\frac{1}{7}-\frac{1}{16}+\frac{1}{16}-\frac{1}{31}+\frac{1}{31}-\frac{1}{43}+\frac{1}{43}-\frac{1}{49}\right)\)

\(=\frac{13}{3}\left(\frac{1}{7}-\frac{1}{49}\right)\) 

\(\Rightarrow\frac{A}{B}=\frac{\frac{17}{3}\left(\frac{1}{7}-\frac{1}{49}\right)}{\frac{13}{3}\left(\frac{1}{7}-\frac{1}{49}\right)}\)\(=\frac{\frac{17}{3}}{\frac{13}{3}}=\frac{17}{13}\)

\(\frac{A}{B}=\frac{17}{13}\)

Chúc bạn học tốt !!!!