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\(E=\frac{2}{3.5}+\frac{7}{5.12}+\frac{9}{4.39}=\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{12}+\frac{27}{12.39}=\frac{1}{3}-\frac{1}{12}+\frac{1}{12}-\frac{1}{39}=\frac{1}{3}-\frac{1}{39}=\frac{4}{13}\)
\(B=\frac{7}{3.13}+\frac{7}{13.23}+...+\frac{7}{53.63}\)
\(B=10.\left(\frac{1}{3.13}+\frac{1}{13.23}+....+\frac{1}{53.63}\right)\)
\(B=10.\left(\frac{1}{3}-\frac{1}{13}+\frac{1}{13}-\frac{1}{23}+...+\frac{1}{53}+\frac{1}{63}\right)\)
\(B=10.\left(\frac{1}{3}-\frac{1}{63}\right)\)
\(B=10.\frac{20}{63}\)
\(B=\frac{200}{63}\)
\(\frac{7}{3.13}+\frac{7}{13.23}+\frac{7}{23.33}+\frac{7}{33.43}\)
\(=\frac{7}{10}\left(\frac{10}{3.13}+\frac{10}{13.23}+\frac{10}{23.33}+\frac{10}{33.43}\right)\)
\(=\frac{7}{10}\left(\frac{1}{3}-\frac{1}{13}+\frac{1}{13}-\frac{1}{23}+\frac{1}{23}-\frac{1}{33}+\frac{1}{33}-\frac{1}{43}\right)\)
\(=\frac{7}{10}\left(\frac{1}{3}-\frac{1}{43}\right)\)
\(=\frac{7}{10}\left(\frac{43}{129}-\frac{3}{129}\right)\)
\(=\frac{7}{10}.\frac{40}{129}\)
\(=\frac{28}{129}\)
mk làm đúng rồi nha, ko tin bấm thử máy tính
7/3.13 + 7/13.23 + 7/23.33 + 7/33.43
= 7/10.(1/3-1/13+1/13-1/23+1/23-1/33+1/33-1/43)
= 7/10.(1/3-1/43)
= 7/10 . 14/43
= 49/215
phần a dễ bạn tự làm đi tử thì bạn tính như bình thường còn mẫu thì:7.(\(\frac{1}{3.13}\)+\(\frac{1}{13.23}\)+\(\frac{1}{23.33}\))
\(\frac{7}{10}\).(\(\frac{1}{3}\)-\(\frac{1}{33}\))=\(\frac{7}{33}\)
b)(1+1/3+1/5+..+1/199)-(1/2+1/4+...+1/200)
(1+1/2+1/3+...+1/199+1/200)-(1/2+1/2+1/4+1/4+...+1/200+1/200)
=1+1/2+1/3+...+1/199+1/200-(1+1/2+1/3+...+1/100)
=1/101+1/102+...+1/200
\(A=\frac{7}{3.13}+\frac{7}{13.23}+...+\frac{7}{53.63}\)
\(A=7.\left(\frac{1}{3.13}+\frac{1}{13.23}+...+\frac{1}{53.63}\right)\)
\(10A=7.\left(\frac{10}{3.13}+\frac{10}{13.23}+...+\frac{10}{53.63}\right)\)
\(10A=7.\left(\frac{1}{3}-\frac{1}{63}\right)\)
\(10A=7.\frac{20}{63}\)
\(10A=\frac{20}{9}\)
\(A=\frac{20}{9}:10\)
\(A=\frac{20}{9}.\frac{1}{10}\)
\(A=\frac{2}{9}\)
Vậy A=\(\frac{2}{9}\)
Chúc bạn học tốt~
A = \(\frac{7}{3.13}\)\(+\)\(\frac{7}{13.23}\)\(+\)\(\frac{7}{23.33}\)\(+\).......... \(+\)\(\frac{7}{53.63}\)
\(\Rightarrow\)A = 7 . (\(\frac{1}{3.13}\)\(+\)\(\frac{1}{13.23}\)\(+\)\(\frac{1}{23.33}\)\(+\)...... \(\frac{1}{53.63}\))
\(\Rightarrow\)A = \(\frac{1}{10}\). 7 . ( \(\frac{1}{3}\)\(-\)\(\frac{1}{13}\)\(+\)\(\frac{1}{13}\)\(-\)\(\frac{1}{23}\)\(+\)\(\frac{1}{23}\)\(-\)\(\frac{1}{33}\)\(+\)....... \(+\)\(\frac{1}{53}\)\(-\)\(\frac{1}{63}\))
\(\Rightarrow\)A = \(\frac{7}{10}\). ( \(\frac{1}{3}\)\(-\)\(\frac{1}{63}\))
\(\Rightarrow\)A = \(\frac{7}{10}\). \(\frac{20}{63}\)
\(\Rightarrow\)A = \(\frac{2}{9}\)
Chúc các anh em học tốt !!!
A=\(\frac{7}{10}\)*(\(\frac{10}{3\cdot13}\)+\(\frac{10}{13\cdot23}\)+\(\frac{10}{23\cdot33}\)+\(\frac{10}{43\cdot53}\)+\(\frac{10}{53\cdot63}\))
A=\(\frac{7}{10}\)*(\(\frac{1}{3}\)-\(\frac{1}{13}\)+\(\frac{1}{13}\)-\(\frac{1}{23}\)+\(\frac{1}{23}\)-\(\frac{1}{33}\)+\(\frac{1}{43}\)-\(\frac{1}{53}\)+\(\frac{1}{53}\)-\(\frac{1}{63}\))
A=\(\frac{7}{10}\)*(\(\frac{1}{3}\)-\(\frac{1}{33}\)+\(\frac{1}{43}\)-\(\frac{1}{63}\))
A=\(\frac{7}{10}\)*(\(\frac{10}{33}\)+\(\frac{20}{2709}\))
A=\(\frac{7}{10}\)*\(\frac{9250}{29799}\)
A=\(\frac{925}{4257}\)
\(S=\frac{5}{3.13}+\frac{5}{13.23}+.....+\frac{5}{83.93}\)
\(2S=\frac{2.5}{3.13}+\frac{2.5}{13.23}+....+\frac{2.5}{83.93}\)
\(2S=\frac{10}{3.13}+\frac{10}{13.23}+.....+\frac{10}{83.93}\)
\(2S=\frac{1}{3}-\frac{1}{13}+\frac{1}{13}-\frac{1}{23}+....+\frac{1}{83}-\frac{1}{93}\)
\(2S=\frac{1}{3}-\frac{1}{93}=\frac{30}{93}\)
\(S=\frac{30}{93}.\frac{1}{2}=\frac{15}{93}\)
Sửa đề:
\(S=\frac{5}{3.13}+\frac{5}{13.23}+.....+\frac{5}{83.93}\)
\(S=\frac{1}{2}.\left(\frac{1}{3}-\frac{1}{13}+\frac{1}{13}-\frac{1}{23}+....+\frac{1}{83}-\frac{1}{93}\right)\)
\(S=\frac{1}{2}.\left(\frac{1}{3}-\frac{1}{93}\right)\)
\(S=\frac{1}{2}.\left(\frac{31}{93}-\frac{1}{93}\right)\)
\(S=\frac{1}{2}.\frac{10}{31}\)
\(S=\frac{5}{31}\)
\(=\frac{3}{3.13}+\frac{3}{13.23}+...+\frac{3}{1993.2003}\)
\(=\frac{1}{10}.\left(1-\frac{3}{13}+\frac{3}{13}-\frac{3}{23}+...+\frac{3}{1993}-\frac{3}{2003}\right)\)
\(=\frac{1}{10}.\left(1-\frac{3}{2003}\right)\)
\(=\frac{1}{10}.\frac{2000}{2003}\)
\(=\frac{200}{2003}\)
Đặt \(A=\frac{1}{13}+\frac{3}{13.23}+\frac{3}{23.33}+...+\frac{3}{1993.2003}\)
\(\Rightarrow A=\frac{3}{3.13}+\frac{3}{13.23}+\frac{3}{23.33}+...+\frac{3}{1993.2003}\)
\(\Rightarrow A=3\left(\frac{1}{3.13}+\frac{1}{13.23}+\frac{1}{23.33}+...+\frac{1}{1993.2003}\right)\)
\(\Rightarrow A=\frac{3}{10}\left(\frac{10}{3.13}+\frac{10}{13.23}+\frac{10}{23.33}+...+\frac{10}{1993.2003}\right)\)
\(\Rightarrow A=\frac{3}{10}\left(\frac{1}{3}-\frac{1}{13}+\frac{1}{13}-\frac{1}{23}+\frac{1}{23}-\frac{1}{33}+...+\frac{1}{1993}-\frac{1}{2003}\right)\)
\(\Rightarrow A=\frac{3}{10}\left(\frac{1}{3}-\frac{1}{2003}\right)\)
\(\Rightarrow A=\frac{3}{10}.\left(\frac{2003}{6009}-\frac{3}{6009}\right)\)
\(\Rightarrow A=\frac{3}{10}.\frac{2000}{6009}\)
\(\Rightarrow A=\frac{200}{2003}\)