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

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

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

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

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

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

\(VậyA=\frac{2}{55}\)

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

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

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

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

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

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

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

Bạn tham khảo bài làm của mình nhé !!

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

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

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

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

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

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

\(\frac{1}{2}-\frac{1}{3.7}-\frac{1}{7.11}-...-\frac{1}{23.27}=\frac{1}{2}-\left(\frac{1}{3.7}+\frac{1}{7.11}+...+\frac{1}{23.27}\right)=\frac{1}{2}-\frac{1}{4}\cdot\left(\frac{1}{3}-\frac{1}{7}+\frac{1}{7}-\frac{1}{11}+...+\frac{1}{23}-\frac{1}{27}\right)=\frac{1}{2}-\frac{1}{4}\cdot\left(\frac{1}{3}-\frac{1}{27}\right)=\frac{1}{2}-\frac{1}{4}\cdot\frac{8}{27}=\frac{23}{54}\)

Lời giải:

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

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

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

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

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

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

\(\frac{4}{3.7}+\frac{4}{7.11}+\frac{4}{11.15}+...+\frac{4}{23.27}\)

\(=\frac{1}{3}-\frac{1}{7}+\frac{1}{7}-\frac{1}{11}+...+\frac{1}{23}-\frac{1}{27}\)

\(=\frac{1}{3}-\frac{1}{27}+0+0+0+0\)

\(=\frac{8}{27}\)

Ta có : \(\frac{4}{3.7}+\frac{4}{7.11}+\frac{4}{11.15}+\frac{4}{15.19}+\frac{4}{19.23}+\frac{4}{23.27}\)

\(=\frac{1}{3}-\frac{1}{7}+\frac{1}{7}-\frac{1}{11}+.....+\frac{1}{23}-\frac{1}{27}\)

\(=\frac{1}{3}-\frac{1}{27}\)

\(=\frac{8}{27}\)

\(\frac{4}{3.7}+\frac{4}{7.11}+\frac{4}{11.15}+\frac{4}{15.19}+\frac{4}{19.23}+\frac{4}{23.27}\)

\(=\frac{7-3}{3.7}+\frac{11-7}{7.11}+.....+\frac{27-23}{23.27}\)

\(=\frac{1}{3}-\frac{1}{7}+\frac{1}{7}-\frac{1}{11}+....+\frac{1}{23}-\frac{1}{27}\)

\(=\frac{1}{3}-\frac{1}{27}=\frac{8}{27}\)

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

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

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

=1/2*4/55

=2/55

\(\frac{1}{2}-\frac{1}{3.7}-\frac{1}{7.11}-\frac{1}{11.15}-\frac{1}{15.19}-\frac{1}{19.23}-\frac{1}{23.27}\)

\(=\frac{1}{2}-\left(\frac{1}{3.7}+\frac{1}{7.11}+\frac{1}{11.15}+...+\frac{1}{23.27}\right)\)

\(=\frac{1}{2}-\frac{1}{4}.\left(\frac{1}{3}-\frac{1}{7}+\frac{1}{7}-\frac{1}{11}+...+\frac{1}{23}-\frac{1}{27}\right)\)

\(=\frac{1}{2}-\frac{1}{4}.\left(\frac{1}{3}-\frac{1}{27}\right)\)

\(=\frac{1}{2}-\frac{1}{4}.\frac{8}{27}\)

\(=\frac{23}{54}\)

                                     Ta có : 

                            \(\frac{1}{2}-\frac{1}{3.7}-\frac{1}{7.11}-\frac{1}{11.15}-\frac{1}{15.19}-\frac{1}{19.23}-\frac{1}{23.27}\)

                          \(=\frac{1}{2}-\left(\frac{1}{3.7}+\frac{1}{7.11}+\frac{1}{11.15}+\frac{1}{15.19}+\frac{1}{19.23}+\frac{1}{23.27}\right)\)

                           \(=\frac{1}{2}-\frac{1}{4}.\left(\frac{4}{3.7}+\frac{4}{7.11}+\frac{4}{11.15}+\frac{4}{15.19}+\frac{4}{19.23}+\frac{4}{23.27}\right)\)

                         \(=\frac{1}{2}-\frac{1}{4}.\left(\frac{1}{3}-\frac{1}{7}+\frac{1}{7}-\frac{1}{11}+\frac{1}{11}-\frac{1}{15}+\frac{1}{15}-\frac{1}{19}+\frac{1}{19}-\frac{1}{23}+\frac{1}{23}-\frac{1}{27}\right)\)

                           \(=\frac{1}{2}-\frac{1}{4}.\left(\frac{1}{3}-\frac{1}{27}\right)\)

                            \(=\frac{1}{2}-\frac{1}{4}.\frac{8}{27}\)

                             \(=\frac{1}{2}-\frac{2}{27}=\frac{27-4}{54}=\frac{23}{54}\)

                              Ủng hộ mk nha !!! ^_^

thanks