TH1) 

\(B=\dfrac{2}{11x15}+\dfrac{2}{15x19}+\dfrac{2}{19x23}+......+\dfrac{2}{51x55}\)

\(B=\dfrac{2}{11}-\dfrac{2}{15}+\dfrac{2}{15}-\dfrac{2}{19}+\dfrac{2}{19}-\dfrac{2}{23}+.....+\dfrac{2}{51}-\dfrac{2}{55}\)

\(B=\dfrac{2}{11}-\dfrac{2}{55}\)

\(B=\dfrac{8}{55}\)

TH2)

\(B=\dfrac{2}{11x15}-\dfrac{2}{15x19}-\dfrac{2}{19x23}-......-\dfrac{2}{51x55}\)

\(B=\dfrac{2}{11}+\dfrac{2}{15}-\dfrac{2}{15}+\dfrac{2}{19}-\dfrac{2}{19}+\dfrac{2}{23}-....-\dfrac{2}{51}+\dfrac{2}{55}\)

\(B=\dfrac{2}{11}+\dfrac{2}{55}\)

\(B=\dfrac{12}{55}\)

B=2/11x15+2/15x19+2/19x23+...+2/51x55

=>2B=4/11x15+4/15x19+4/19x23+...+4/51x55

=1/11-1/15+1/15-1/19+...+1/51-1/55

=1/11-1/55

=5/55-1/55

=4/55

=>B=4/55 : 2

=4/55.1/2

=2/55

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

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

\(C=\frac{2}{11.15}+\frac{2}{15.19}+\frac{2}{19.13}+...+\frac{2}{51.55}\)

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

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

\(2.C=\frac{1}{11}-\frac{1}{55}\)

\(2.C=\frac{5}{55}-\frac{1}{55}=\frac{4}{55}\)

\(C=\frac{4}{55}:2=\frac{4}{55}.\frac{1}{2}=\frac{2}{55}\)

Vậy \(C=\frac{2}{55}\)

đọc kĩ đi bạn ơi

có cả số 1 nữa

a)\(A=\frac{2}{3}+\frac{2}{6}+\frac{2}{12}+\frac{2}{24}+\frac{2}{48}+\frac{2}{96}+\frac{2}{192}\)

\(\frac{1}{2}xA=\frac{1}{3}+\frac{1}{6}+\frac{1}{12}+\frac{1}{24}+\frac{1}{48}+\frac{1}{96}+\frac{1}{192}\)

\(\frac{1}{4}xA=\frac{1}{6}+\frac{1}{12}+\frac{1}{24}+\frac{1}{48}+\frac{1}{96}+\frac{1}{192}+\frac{1}{384}\)

\(\frac{1}{4}xA-\frac{1}{2}xA=\frac{1}{3}-\frac{1}{384}\)

\(\frac{1}{4}xA=\frac{127}{384}\)

\(A=\frac{127}{384}:\frac{1}{4}\)

\(A=\frac{127}{96}\)

\(\frac{1}{2}+\frac{5}{6}+\frac{11}{12}+\frac{19}{20}+...+\frac{89}{90}\)

\(=1-\frac{1}{2}+1-\frac{1}{6}+1-\frac{1}{12}+1-\frac{1}{20}+...+1-\frac{1}{90}\)

\(=9-\left(\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+...+\frac{1}{90}\right)\)

\(=9-\left(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{9.10}\right)\)

\(=9-\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+....+\frac{1}{9}-\frac{1}{10}\right)\)

\(=9-\left(1-\frac{1}{10}\right)\)

\(=9-\frac{9}{10}=\frac{81}{10}\)

\(A=\dfrac{4}{3.7}+\dfrac{4}{7.11}+\dfrac{4}{11.15}+\dfrac{4}{15.19}+\dfrac{4}{19.23}+\dfrac{4}{23.27}\)(Dấu . là dấu nhân)
\(=\dfrac{1}{3}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{11}+\dfrac{1}{11}-\dfrac{1}{15}+\dfrac{1}{15}-\dfrac{1}{19}+\dfrac{1}{19}-\dfrac{1}{23}+\dfrac{1}{23}-\dfrac{1}{27}\)
\(=\dfrac{1}{3}-\dfrac{1}{27}\)
\(=\dfrac{9}{27}-\dfrac{1}{27}\)
\(=\dfrac{8}{27}\)

A = 4/3x7 + 4/7x11+ 4/11x15 + 4/15x19 + 4/19 x23 + 4/23 x 27

A = 1/3-1/7+1/7-1/11+1/11-1/15+1/15-1/19+1/19-1/23+1/23 -1/27

A = 1/3 - 1/27

A = 8/27

SỬa đề: \(\dfrac{4}{3\cdot7}+\dfrac{4}{7\cdot11}+...+\dfrac{4}{23\cdot27}\)

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

=1/3-1/27

=8/27

Ta có :

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

\(=\frac{1}{3}-\frac{1}{7}+\frac{1}{7}-\frac{1}{11}+\frac{1}{11}-\frac{1}{15}+....+\frac{1}{23}-\frac{1}{27}\)

\(=\frac{1}{3}-\frac{1}{27}==\frac{9}{27}-\frac{1}{27}=\frac{8}{27}\)

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

\(A=\frac{1}{3}-\frac{1}{7}+\frac{1}{7}-\frac{1}{11}+\frac{1}{11}-\frac{1}{15}+...+\frac{1}{23}-\frac{1}{27}\)

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

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

\(\frac{4}{3\cdot7}+\frac{4}{7\cdot11}+\frac{4}{11\cdot15}+\frac{4}{15\cdot19}+\frac{4}{19\cdot23}+\frac{4}{23.27}\)

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

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

\(=\frac{4}{4}\cdot\frac{8}{27}=\frac{8}{27}\)