Ta có : \(A=\frac{40}{31.39}+\frac{35}{39.46}+\frac{30}{46.52}+\frac{25}{52.57}=5\left(\frac{8}{31.39}+\frac{7}{39.46}+\frac{6}{46.52}+\frac{5}{52.57}\right)\)

\(=5\left(\frac{1}{31}-\frac{1}{39}+\frac{1}{39}-\frac{1}{46}+\frac{1}{46}-\frac{1}{52}+\frac{1}{52}-\frac{1}{57}\right)=5\left(\frac{1}{31}-\frac{1}{57}\right)=5.\frac{26}{1767}=\frac{130}{1767}\)

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

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

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

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

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

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

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

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

haizzz ghi đề lại đàng hoàng đi cha nội ơi

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