A=12/15 + 28/315

            A=8/9 

\(\frac{1}{15}+\frac{1}{35}+\frac{1}{63}+...+\frac{1}{\left(2x+1\right)\left(2x+3\right)}=\frac{15}{93}\)

\(\frac{1}{3.5}+\frac{1}{5.7}+\frac{1}{7.9}+...+\frac{1}{\left(2x+1\right)\left(2x+3\right)}=\frac{15}{93}\)

\(\frac{1}{2}\left(\frac{2}{3.5}+\frac{2}{5.7}+\frac{2}{7.9}+...+\frac{2}{\left(2x+1\right)\left(2x+3\right)}\right)=\frac{15}{93}\)

\(\frac{1}{2}\)\(\left(\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{2x+1}-\frac{1}{2x+3}\right)\)\(=\frac{15}{93}\)

\(\frac{1}{2}\left(\frac{1}{3}-\frac{1}{2x+3}\right)=\frac{15}{93}\)

\(\frac{1}{3}-\frac{1}{2x+3}=\frac{15}{93}:\frac{1}{2}=\frac{10}{31}\)

\(\frac{1}{2x+3}=\frac{1}{3}-\frac{10}{31}=\frac{1}{93}\)

\(\Rightarrow2x+3=93\rightarrow2x=90\rightarrow x=45\)

Đề sai rồi em, mẫu số đều là số lẻ thì 120 ko theo quy luật 

Các hạng trong S đều là số lẻ mà 120 là số chẵn nên đề sai nhé

11 x 13 = 143. 120 ở đâu ra?

\(A=\dfrac{1}{3}+\dfrac{1}{15}+\dfrac{1}{35}+\dfrac{1}{63}+...+\dfrac{1}{899}\\ 2A=2\cdot\dfrac{1}{3}+2\cdot\dfrac{1}{15}+2\cdot\dfrac{1}{35}+2\cdot\dfrac{1}{63}+...+2\cdot\dfrac{1}{899}\\ 2A=1-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+...+\dfrac{1}{29}-\dfrac{1}{31}\\ 2A=1-\dfrac{1}{31}\\ 2A=\dfrac{30}{31}\\ A=\dfrac{30}{31}\div2\\ A=\dfrac{30}{31\cdot2}=\dfrac{15}{31}\)

:))

Đặt phép tính cần tìm là A

\(A=\dfrac{1}{1.3}+\dfrac{1}{3.5}+\dfrac{1}{5.7}+\dfrac{1}{7.9}+\dfrac{1}{9.11}+\dfrac{1}{11.13}\)

\(2A=\dfrac{2}{1.3}+\dfrac{2}{3.5}+\dfrac{2}{5.7}+\dfrac{2}{7.9}+\dfrac{2}{9.11}+\dfrac{2}{11.13}\)

\(2A=1-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{9}+\dfrac{1}{9}-\dfrac{1}{11}+\dfrac{1}{11}-\dfrac{1}{13}\)

\(2A=1-\dfrac{1}{13}\)

\(2A=\dfrac{12}{13}\)

\(A=\dfrac{6}{13}\)

\(A=\dfrac{1}{3}+\dfrac{1}{15}+...+\dfrac{1}{143}\\ =\dfrac{1}{1\times3}+\dfrac{1}{3\times5}+...+\dfrac{1}{11\times13}\\ =\dfrac{1}{2}\times\left(\dfrac{1}{1\times3}+\dfrac{1}{3\times5}+...+\dfrac{1}{11\times13}\right)\\ =\dfrac{1}{2}\times\left(1-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+...+\dfrac{1}{11}-\dfrac{1}{13}\right)\\ =\dfrac{1}{2}\times\dfrac{12}{13}\\ =\dfrac{6}{13}\)

\(\Leftrightarrow\dfrac{4}{9}:x=\dfrac{8}{3}\)

hay \(x=\dfrac{4}{9}\cdot\dfrac{3}{8}=\dfrac{1}{2}\cdot\dfrac{1}{3}=\dfrac{1}{6}\)

cái này tính cái gì thế

ko hiểu

\(C=\frac{1}{3}+\frac{1}{15}+\frac{1}{35}+...+\frac{1}{1023}\)

\(=\frac{1}{1\cdot3}+\frac{1}{3\cdot5}+\frac{1}{5\cdot7}....+\frac{1}{31\cdot33}\)

\(=\frac{1}{2}\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+...+\frac{1}{31}-\frac{1}{33}\right)\)

\(=\frac{1}{2}\left(1-\frac{1}{33}\right)\)

\(=\frac{1}{2}\cdot\frac{32}{33}\)

\(=\frac{32}{66}=\frac{16}{33}\)

Vậy \(A=\frac{16}{33}\)

HOK TỐT .

\(C=\frac{1}{3}+\frac{1}{15}+\frac{1}{35}+...+\frac{1}{1023}\)

\(C=\frac{1}{1\cdot3}+\frac{1}{3\cdot5}+\frac{1}{5\cdot7}+...+\frac{1}{31\cdot33}\)

\(C=\frac{1}{2}\left(\frac{2}{1\cdot3}+\frac{2}{3\cdot5}+\frac{2}{5\cdot7}+...+\frac{2}{31\cdot33}\right)\)

\(C=\frac{1}{2}\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{31}-\frac{1}{33}\right)\)

\(C=\frac{1}{2}\left(1-\frac{1}{33}\right)\)

\(C=\frac{1}{2}\cdot\frac{32}{33}\)

\(C=\frac{16}{33}\)

\(C=\frac{1}{3}+\frac{1}{15}+\frac{1}{35}+\frac{1}{63}+...+\frac{1}{1023}\)

\(=\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+...+\frac{1}{31.33}\)

\(=\frac{1}{2}\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+...+\frac{1}{31}-\frac{1}{33}\right)\)

\(=\frac{1}{2}\left(1-\frac{1}{33}\right)\)

\(=\frac{1}{2}.\frac{32}{33}\)

\(=\frac{16}{33}\)

#Chúc bạn học tốt#