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anh ơi anh anh dẹp cho em nhờ

bạn viết sai đề rồi . phải là 7 . 9 chứ

Đặt \(A=\frac{1}{2.9}+\frac{1}{9.7}+\frac{1}{7.19}+...+\frac{1}{252.509}\)

\(\Leftrightarrow A=\frac{2}{5}.\left(\frac{5}{4.9}+\frac{5}{9.14}+\frac{5}{14.19}+...+\frac{5}{504.509}\right)\)

\(\Leftrightarrow A=\frac{2}{5}.\left(\frac{1}{4}-\frac{1}{9}+\frac{1}{9}-\frac{1}{14}+\frac{1}{14}-\frac{1}{19}+...+\frac{1}{504}-\frac{1}{509}\right)\)

\(\Leftrightarrow A=\frac{2}{5}.\left(\frac{1}{4}-\frac{1}{509}\right)\)

\(\Leftrightarrow A=\frac{2}{5}.\frac{505}{2036}\)

\(\Leftrightarrow A=\frac{101}{1018}\)

~ Hok tốt ~

#)Giải :

\(A=\frac{1}{2.9}+\frac{1}{9.7}+\frac{1}{7.19}+...+\frac{1}{252.509}\)

\(A=\frac{2}{5}\left(\frac{5}{4.9}+\frac{5}{9.14}+\frac{5}{14.19}+...+\frac{5}{504.509}\right)\)

\(A=\frac{2}{5}\left(\frac{1}{4}-\frac{1}{9}+\frac{1}{9}-\frac{1}{14}+\frac{1}{14}-\frac{1}{17}+...+\frac{1}{504}-\frac{1}{509}\right)\)

\(A=\frac{2}{5}\left(\frac{1}{4}-\frac{1}{509}\right)\)

\(A=\frac{2}{5}\times\frac{505}{2036}\)

\(A=\frac{101}{1018}\)

\(A=\frac{1}{7}\left[\frac{1}{2}-\frac{1}{9}+...+\frac{1}{252}-\frac{1}{509}\right]\)

\(A=\frac{1}{7}.\left[\frac{1}{2}-\frac{1}{509}\right]\)

\(A=\frac{1}{7}.\left[\frac{507}{1018}\right]=\frac{507}{7126}\)

mk nghĩ là vậy đó, ủng hộ mk nha

A =$\genfrac{}{}{0ex}{}{1}{2.9}+\genfrac{}{}{0ex}{}{1}{9.7}+\genfrac{}{}{0ex}{}{1}{7.19}+...+\genfrac{}{}{0ex}{}{1}{252.509}$

A = 2. ($\genfrac{}{}{0ex}{}{1}{4.9}+\genfrac{}{}{0ex}{}{1}{9.14}+\genfrac{}{}{0ex}{}{1}{14.19}+...+\genfrac{}{}{0ex}{}{1}{504.509}$)

A =$\genfrac{}{}{0ex}{}{2}{5}$($\genfrac{}{}{0ex}{}{1}{4}-\genfrac{}{}{0ex}{}{1}{9}+\genfrac{}{}{0ex}{}{1}{9}-\genfrac{}{}{0ex}{}{1}{14}+\genfrac{}{}{0ex}{}{1}{14}-\genfrac{}{}{0ex}{}{1}{19}+...+\genfrac{}{}{0ex}{}{1}{504}-\genfrac{}{}{0ex}{}{1}{509}$)

A =$\genfrac{}{}{0ex}{}{2}{5}$($\genfrac{}{}{0ex}{}{1}{4}-\genfrac{}{}{0ex}{}{1}{509}$)

A =$\genfrac{}{}{0ex}{}{2}{5}$($\genfrac{}{}{0ex}{}{509}{2036}-\genfrac{}{}{0ex}{}{4}{2036}$)

A =$\genfrac{}{}{0ex}{}{2}{5}$.$\genfrac{}{}{0ex}{}{505}{2036}$

A =$\genfrac{}{}{0ex}{}{101}{1018}$

\(A=\frac{2}{4.9}+\frac{2}{9.14}+\frac{2}{14.19}+...+\frac{2}{504.509}\)

\(A=\frac{2}{5}\left(\frac{1}{4}-\frac{1}{9}+\frac{1}{9}-\frac{1}{14}+...+\frac{1}{504}-\frac{1}{509}\right)\)

\(A=\frac{2}{5}\left(\frac{1}{4}-\frac{1}{509}\right)=...\)

\(B=\frac{1.4+1}{1.4}+\frac{4.7+1}{4.7}+\frac{7.10+1}{7.10}+...+\frac{100.103+1}{100.103}\)

\(B=1+\frac{1}{1.4}+1+\frac{1}{4.7}+...+1+\frac{1}{100.103}\)

\(B=34+\frac{1}{3}\left(\frac{3}{1.4}+\frac{3}{4.7}+...+\frac{3}{100.103}\right)\)

\(B=34+\frac{1}{3}\left(1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+...+\frac{1}{100}-\frac{1}{103}\right)\)

\(B=34+\frac{1}{3}\left(1-\frac{1}{103}\right)=...\)