\(A=5\left(\frac{8}{31.39}+\frac{7}{39.46}+\frac{6}{46.52}+\frac{5}{52.57}\right)=8\left(\frac{1}{31}-\frac{1}{39}+\frac{1}{39}-\frac{1}{46}+\frac{1}{46}-\frac{1}{52}+\frac{1}{52}-\frac{1}{57}\right)\)=\(8\left(\frac{1}{31}-\frac{1}{57}\right)=8\left(\frac{26}{1767}\right)=\frac{208}{1767}\)

208/1767

tich tớ nhé Nguyễn Trọng Đức Siu Đệp Choai

\(A=\frac{40}{31}-\frac{40}{39}+\frac{35}{39}-\frac{35}{46}+\frac{30}{46}-\frac{30}{52}+\frac{25}{52}-\frac{25}{57}\)

\(A=\frac{1}{31}-\frac{1}{39}+\frac{1}{39}-\frac{1}{46}+\frac{1}{46}-\frac{1}{52}+\frac{1}{52}-\frac{1}{57}\)

\(A=\frac{1}{31}-\frac{1}{57}\)

tiếp theo tính A và B tương tự rồi tìm tỉ số, ta lấy A:B*100=?%

Bg (sửa phân số cuối thành \(\frac{20}{57.61}\)

Ta có: C = \(\frac{40}{31.39}+\frac{35}{39.46}+\frac{30}{46.52}+\frac{25}{52.57}+\frac{20}{57.61}\)

=> C = \(\frac{5.8}{31.39}+\frac{5.7}{39.46}+\frac{5.6}{46.52}+\frac{5.5}{52.57}+\frac{5.4}{57.61}\)

=> C = \(5.\left(\frac{8}{31.39}+\frac{7}{39.46}+\frac{6}{46.52}+\frac{5}{52.57}+\frac{4}{57.61}\right)\)

=> C = \(5.\left(\frac{1}{31}-\frac{1}{39}+\frac{1}{39}-\frac{1}{46}+\frac{1}{46}-\frac{1}{52}+\frac{1}{52}-\frac{1}{57}+\frac{1}{57}-\frac{1}{61}\right)\)

=> C = \(5.\left(\frac{1}{31}-\frac{1}{61}\right)\)

=> C = \(5.\frac{30}{1891}\)  (bắt đầu xấu)

=> C = \(\frac{150}{1891}\)(so ugly)

Vậy C = \(\frac{150}{1891}\)(kết thúc không có hậu)

Đăt \(A=\dfrac{40}{31.39}+\dfrac{35}{39.46}+\dfrac{30}{46.52}+\dfrac{25}{52.57}+\dfrac{20}{57.61}\)

\(\Leftrightarrow A=5.\left(\dfrac{8}{31.39}+\dfrac{7}{39.46}+\dfrac{6}{46.52}+\dfrac{5}{52.57}+\dfrac{4}{57.61}\right)\)

\(\Leftrightarrow A=5.\left(\dfrac{1}{31}-\dfrac{1}{39}+\dfrac{1}{39}-\dfrac{1}{46}+\dfrac{1}{46}-\dfrac{1}{52}+\dfrac{1}{52}-\dfrac{1}{57}+\dfrac{1}{57}-\dfrac{1}{61}\right)\)\(\Leftrightarrow A=5.\left(\dfrac{1}{31}-\dfrac{1}{61}\right)\)

\(\Leftrightarrow A=5.\dfrac{30}{31.61}\)

\(\Leftrightarrow A=\dfrac{150}{1891}\)

Ta có: \(A=\frac{2}{1\cdot5}+\frac{3}{5\cdot11}+\frac{4}{11\cdot19}+\frac{5}{19\cdot29}+\frac{6}{29\cdot41}\)

\(=\frac12\left(\frac{4}{1\cdot5}+\frac{6}{5\cdot11}+\frac{8}{11\cdot19}+\frac{10}{19\cdot29}+\frac{12}{29\cdot41}\right)\)

\(=\frac12\left(1-\frac15+\frac15-\frac{1}{11}+\frac{1}{11}-\frac{1}{19}+\frac{1}{19}-\frac{1}{29}+\frac{1}{29}-\frac{1}{41}\right)\)

\(=\frac12\left(1-\frac{1}{41}\right)=\frac12\cdot\frac{40}{41}=\frac{20}{41}\)

Ta có: \(B=\frac{40}{31\cdot39}+\frac{35}{39\cdot46}+\frac{30}{46\cdot52}+\frac{25}{52\cdot57}+\frac{20}{57\cdot61}\)

\(=5\left(\frac{8}{31\cdot39}+\frac{7}{39\cdot46}+\frac{6}{46\cdot52}+\frac{5}{52\cdot57}+\frac{4}{57\cdot61}\right)\)

\(=5\left(\frac{1}{31}-\frac{1}{39}+\frac{1}{39}-\frac{1}{46}+\frac{1}{46}-\frac{1}{52}+\frac{1}{52}-\frac{1}{57}+\frac{1}{57}-\frac{1}{61}\right)\)

\(=5\left(\frac{1}{31}-\frac{1}{61}\right)=5\cdot\frac{30}{61\cdot31}=\frac{150}{1891}\)

\(\frac{A}{B}=\frac{20}{41}:\frac{150}{1891}=\frac{20}{41}\cdot\frac{1891}{150}=\frac{2}{15}\cdot\frac{1891}{41}=\frac{3782}{615}\)