Đặt :

\(A=\dfrac{2}{11.15}+\dfrac{2}{15.19}+.......+\dfrac{2}{51.55}\)

\(\Leftrightarrow2A=\dfrac{4}{11.15}+\dfrac{4}{15.19}+.........+\dfrac{4}{51.55}\)

\(\Leftrightarrow2A=\dfrac{1}{11}-\dfrac{1}{15}+\dfrac{1}{15}-\dfrac{1}{19}+......+\dfrac{1}{51}-\dfrac{1}{55}\)

\(\Leftrightarrow2A=\dfrac{1}{11}-\dfrac{1}{55}\)

\(\Leftrightarrow2A=\dfrac{4}{55}\)

\(\Leftrightarrow A=\dfrac{4}{110}\)

\(A=\frac{1}{2}.\left(\frac{4}{11.15}+\frac{4}{15.19}+\frac{4}{19.23}+...+\frac{4}{51.55}\right)\)

\(=\frac{1}{2}.\left(\frac{1}{11}-\frac{1}{15}+\frac{1}{15}-\frac{1}{19}+\frac{1}{19}-\frac{1}{23}+...+\frac{1}{51}-\frac{1}{55}\right)\)

\(=\frac{1}{2}.\left(\frac{1}{11}-\frac{1}{55}\right)=\frac{1}{2}.\frac{4}{55}=\frac{2}{55}\)

\(\Rightarrow A=\frac{2}{4}\left(\frac{1}{11.15}+\frac{1}{15.19}+\frac{1}{19.23}+.....+\frac{1}{51.55}\right)\)

\(\Rightarrow A=\frac{1}{2}\left(\frac{1}{11}-\frac{1}{15}+\frac{1}{15}-\frac{1}{19}+...+\frac{1}{51}-\frac{1}{55}\right)\)

\(\Rightarrow A=\frac{1}{2}\left(\frac{1}{11}-\frac{1}{55}\right)=\frac{1}{2}.\frac{4}{55}=\frac{2}{55}\)

\(VậyA=\frac{2}{55}\)

Bạn tham khảo bài làm của mình nhé !!

\(A=\frac{2}{11.15}+\frac{2}{15.19}+\frac{2}{19.23}...+\frac{2}{51.55}\)

\(\Leftrightarrow2A=\frac{4}{11.15}+\frac{4}{15.19}+\frac{4}{19.23}+...+\frac{4}{51.55}\)

\(\Leftrightarrow2A=\frac{1}{11}-\frac{1}{15}+\frac{1}{15}-\frac{1}{19}+...+\frac{1}{51}-\frac{1}{55}\)

\(\Leftrightarrow2A=\frac{1}{11}-\frac{1}{55}\)

\(\Leftrightarrow2A=\frac{4}{55}\)

\(\Leftrightarrow A=\frac{4}{110}\)

\(A=\dfrac{-5}{3}\cdot\dfrac{11}{2}\cdot\dfrac{4}{3}=\dfrac{-20\cdot11}{2\cdot9}=\dfrac{-110}{9}\)

\(B=\dfrac{2}{4}\left(\dfrac{4}{11\cdot15}+\dfrac{4}{15\cdot19}+...+\dfrac{4}{51\cdot55}\right)\)

\(=\dfrac{1}{2}\left(\dfrac{1}{11}-\dfrac{1}{15}+\dfrac{1}{15}-\dfrac{1}{19}+...+\dfrac{1}{51}-\dfrac{1}{55}\right)\)

=1/2*4/55

=2/55

Lời giải:

\(A=\frac{2}{11.15}+\frac{2}{15.19}+\frac{2}{19.23}+...+\frac{2}{51.55}\)

\(\Rightarrow 2A=\frac{4}{11.15}+\frac{4}{15.19}+\frac{4}{19.23}+...+\frac{4}{51.55}\)

\(=\frac{15-11}{11.15}+\frac{19-15}{15.19}+\frac{23-19}{19.23}+....+\frac{55-51}{51.55}\)

\(=\frac{1}{11}-\frac{1}{15}+\frac{1}{15}-\frac{1}{19}+\frac{1}{19}-\frac{1}{23}+...+\frac{1}{51}-\frac{1}{55}\)

\(=\frac{1}{11}-\frac{1}{55}=\frac{4}{55}\)

\(\Rightarrow A=\frac{2}{55}\)

a) \(A=\frac{2}{11.15}+\frac{2}{15.19}+...+\frac{2}{51.55}\)

\(=\frac{1}{2}\left(\frac{4}{11.15}+\frac{4}{15.19}+...+\frac{4}{51.55}\right)\)

\(=\frac{1}{2}\left(\frac{15-11}{11.15}+\frac{19-15}{15.19}+...+\frac{55-51}{51.55}\right)\)

\(=\frac{1}{2}\left(\frac{1}{11}-\frac{1}{15}+\frac{1}{15}-\frac{1}{19}+...+\frac{1}{51}-\frac{1}{55}\right)\)

\(=\frac{1}{2}\left(\frac{1}{11}-\frac{1}{55}\right)=\frac{2}{55}\)

b) \(\overline{abcabc}=\overline{abc}.1001=\overline{abc}.7.11.13\)suy ra đpcm. 

\(\overline{abcabc}=1001.\overline{abc}=7.11.13.\overline{abc}\)

7, 11, 13 là các số nguyên tố

=\(1\left(\frac{1}{14.15}+\frac{1}{15.19}+......+\frac{1}{51.55}\right)\)

=\(1\left(\frac{1}{14}-\frac{1}{15}\right)+\left(\frac{1}{15}-\frac{1}{19}\right).....+\left(\frac{1}{51}-\frac{1}{55}\right)\)

=\(1\left(\frac{1}{14}-\frac{1}{15}+\frac{1}{15}-\frac{1}{19}....+\frac{1}{51}-\frac{1}{55}\right)\)

=\(1\left(\frac{1}{14}-\frac{1}{55}\right)\)

=\(1.\frac{41}{770}\)

=\(\frac{41}{770}\)