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\(=\frac{3}{3.13}+\frac{3}{13.23}+...+\frac{3}{1993.2003}\)
\(=\frac{1}{10}.\left(1-\frac{3}{13}+\frac{3}{13}-\frac{3}{23}+...+\frac{3}{1993}-\frac{3}{2003}\right)\)
\(=\frac{1}{10}.\left(1-\frac{3}{2003}\right)\)
\(=\frac{1}{10}.\frac{2000}{2003}\)
\(=\frac{200}{2003}\)
Đặt \(A=\frac{1}{13}+\frac{3}{13.23}+\frac{3}{23.33}+...+\frac{3}{1993.2003}\)
\(\Rightarrow A=\frac{3}{3.13}+\frac{3}{13.23}+\frac{3}{23.33}+...+\frac{3}{1993.2003}\)
\(\Rightarrow A=3\left(\frac{1}{3.13}+\frac{1}{13.23}+\frac{1}{23.33}+...+\frac{1}{1993.2003}\right)\)
\(\Rightarrow A=\frac{3}{10}\left(\frac{10}{3.13}+\frac{10}{13.23}+\frac{10}{23.33}+...+\frac{10}{1993.2003}\right)\)
\(\Rightarrow A=\frac{3}{10}\left(\frac{1}{3}-\frac{1}{13}+\frac{1}{13}-\frac{1}{23}+\frac{1}{23}-\frac{1}{33}+...+\frac{1}{1993}-\frac{1}{2003}\right)\)
\(\Rightarrow A=\frac{3}{10}\left(\frac{1}{3}-\frac{1}{2003}\right)\)
\(\Rightarrow A=\frac{3}{10}.\left(\frac{2003}{6009}-\frac{3}{6009}\right)\)
\(\Rightarrow A=\frac{3}{10}.\frac{2000}{6009}\)
\(\Rightarrow A=\frac{200}{2003}\)
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
\(B=\frac{7}{3.13}+\frac{7}{13.23}+...+\frac{7}{53.63}\)
\(B=10.\left(\frac{1}{3.13}+\frac{1}{13.23}+....+\frac{1}{53.63}\right)\)
\(B=10.\left(\frac{1}{3}-\frac{1}{13}+\frac{1}{13}-\frac{1}{23}+...+\frac{1}{53}+\frac{1}{63}\right)\)
\(B=10.\left(\frac{1}{3}-\frac{1}{63}\right)\)
\(B=10.\frac{20}{63}\)
\(B=\frac{200}{63}\)
\(\frac{1}{2003.2002}-\frac{1}{2002.2001}-...-\frac{1}{2.1}\)
\(=\frac{1}{2003.2002}-\left(\frac{1}{1.2}+\frac{1}{3.2}+...+\frac{1}{2001.2002}\right)\)
\(=\frac{1}{2003.2002}-\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-...+\frac{1}{2000}-\frac{1}{2001}+\frac{1}{2001}-\frac{1}{2002}\right)\)
\(=\frac{1}{2003.2002}-\left(1-\frac{1}{2002}\right)\)
\(=\frac{1}{2003.2002}-\frac{2001}{2002}\)
5/3.13 + 5/13.23 + 5/23,33
= 1,975373039 + 5/23,33
=2,18968937
\(E=\frac{2}{3.5}+\frac{7}{5.12}+\frac{9}{4.39}=\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{12}+\frac{27}{12.39}=\frac{1}{3}-\frac{1}{12}+\frac{1}{12}-\frac{1}{39}=\frac{1}{3}-\frac{1}{39}=\frac{4}{13}\)