\(\dfrac{3}{17}A=\dfrac{6}{7.13}+\dfrac{9}{13.22}+\dfrac{15}{22.37}+\dfrac{12}{37.49}\)

\(\dfrac{3}{17}A=\dfrac{1}{7}-\dfrac{1}{13}+\dfrac{1}{13}-\dfrac{1}{22}+\dfrac{1}{22}-\dfrac{1}{37}+\dfrac{1}{37}-\dfrac{1}{49}\)

\(\dfrac{3}{17}A=\dfrac{1}{7}-\dfrac{1}{49}=\dfrac{6}{49}\)

\(A=\dfrac{6}{49}:\dfrac{3}{17}=\dfrac{6.17}{49.3}=\dfrac{34}{49}\)

1. Ta có : (\(\dfrac{-3}{8}\))³ < 0
(\(\dfrac{8}{243}\))³ > 0
=> (\(\dfrac{-3}{8}\))³ < (\(\dfrac{8}{243}\))³
@Cuber Việt

\(\left(\dfrac{-3}{8}\right)^3< 0< \left(\dfrac{8}{243}\right)^3\)

Vậy \(\left(\dfrac{-3}{8}\right)^3< \left(\dfrac{8}{243}\right)^3\)

\(A=\dfrac{34}{7\cdot13}+\dfrac{51}{13\cdot22}+\dfrac{85}{22\cdot37}+\dfrac{68}{37\cdot49}\\ =\dfrac{17}{3}\cdot\dfrac{6}{7\cdot13}+\dfrac{17}{3}\cdot\dfrac{9}{13\cdot22}+\dfrac{17}{3}\cdot\dfrac{15}{22\cdot37}+\dfrac{17}{3}\cdot\dfrac{12}{37\cdot49}\\ =\dfrac{17}{3}\cdot\left(\dfrac{6}{7\cdot13}+\dfrac{9}{13\cdot22}+\dfrac{15}{22\cdot37}+\dfrac{12}{37\cdot49}\right)\\ =\dfrac{17}{3}\cdot\left(\dfrac{1}{7}-\dfrac{1}{13}+\dfrac{1}{13}-\dfrac{1}{22}+\dfrac{1}{22}-\dfrac{1}{37}+\dfrac{1}{37}-\dfrac{1}{49}\right)\\ =\dfrac{17}{3}\cdot\left(\dfrac{1}{7}-\dfrac{1}{49}\right)\\ =\dfrac{17}{3}\cdot\dfrac{6}{49}\\ =\dfrac{34}{49}\)

mk nhầm chỗ câu so sánh phải là mũ 5 nha

Ta có :

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

\(\dfrac{A}{17}=\dfrac{2}{7.13}+\dfrac{3}{13.22}+\dfrac{5}{22.37}+\dfrac{4}{37.49}\)

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

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

\(A.\dfrac{3}{17}=\dfrac{1}{7}-\dfrac{1}{49}\)

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

\(\Rightarrow A=\dfrac{6}{49}:\dfrac{3}{17}=\dfrac{34}{49}\)

\(B=\dfrac{39}{7.16}+\dfrac{65}{16.31}+\dfrac{52}{31.43}+\dfrac{26}{37.49}\)

\(\dfrac{B}{13}=\dfrac{3}{7.16}+\dfrac{5}{16.31}+\dfrac{4}{31.43}+\dfrac{2}{37.49}\)

\(B.\dfrac{3}{13}=\dfrac{9}{7.16}+\dfrac{15}{16.31}+\dfrac{12}{31.43}+\dfrac{6}{43.49}\)

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

\(B.\dfrac{3}{13}=\dfrac{1}{7}-\dfrac{1}{49}\)

\(B.\dfrac{3}{13}=\dfrac{1}{7}-\dfrac{1}{49}=\dfrac{6}{49}\)

\(\Rightarrow B=\dfrac{6}{49}:\dfrac{3}{13}=\dfrac{26}{49}\)

\(\Rightarrow\dfrac{A}{B}=\dfrac{34}{49}:\dfrac{26}{49}=\dfrac{17}{13}\)

Chúc bn học tốt!!!!!!!!!

Link này bạn: Câu hỏi của Quỳnh Anh Shuy - Toán lớp 7 | Học trực tuyến