\(S=\frac{1}{9}+\frac{1}{45}+\frac{1}{105}+\frac{1}{189}+\frac{1}{297}=\frac{1}{3}\left(\frac{1}{3}+\frac{1}{15}+\frac{1}{35}+\frac{1}{63}+\frac{1}{99}\right)\)

\(=\frac{1}{3}.\frac{1}{2}\left(\frac{2}{3}+\frac{2}{15}+\frac{2}{35}+\frac{2}{63}+\frac{2}{99}\right)=\frac{1}{6}\left(\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+\frac{2}{7.9}+\frac{2}{9.11}\right)\)

\(=\frac{1}{6}\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+\frac{1}{9}-\frac{1}{11}\right)=\frac{1}{6}\left(1-\frac{1}{11}\right)=\frac{1}{6}.\frac{10}{11}\)

\(=\frac{5}{33}\)

Bài làm:

\(S=\frac{1}{9}+\frac{1}{45}+\frac{1}{105}+\frac{1}{189}+\frac{1}{297}\)

\(S=\frac{1}{6}\left(\frac{2}{3}+\frac{2}{15}+\frac{2}{35}+\frac{2}{63}+\frac{2}{99}\right)\)

\(S=\frac{1}{6}\left(\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+\frac{2}{7.9}+\frac{2}{9.11}\right)\)

\(S=\frac{1}{6}\left(\frac{3-1}{1.3}+\frac{5-3}{3.5}+\frac{7-5}{5.7}+\frac{9-7}{7.9}+\frac{11-9}{9.11}\right)\)

\(S=\frac{1}{6}\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+\frac{1}{9}-\frac{1}{11}\right)\)

\(S=\frac{1}{6}\left(1-\frac{1}{11}\right)\)

\(S=\frac{1}{6}.\frac{10}{11}=\frac{5}{33}\)

Vậy \(S=\frac{5}{33}\)

Xin lỗi bạn Xyz nhé, mk ko có chép bài bạn đâu! với lại mk cx ko k sai bài bn đâu nhé!

Bạn tải photomath về là giải được ngay ! 

Nhớ k cho mình nhé !

Gọi biểu thức này là A

Ta có :

\(A=\frac{4}{45}+\frac{4}{105}+\frac{4}{189}+\frac{4}{297}+\frac{4}{929}\)

\(\frac{3}{2}A=\frac{3}{2}\times\left(\frac{4}{45}+\frac{4}{105}+\frac{4}{189}+\frac{4}{297}+\frac{4}{929}\right)\)

\(\frac{3}{2}A=\frac{6}{45}+\frac{6}{105}+\frac{6}{189}+\frac{6}{297}+\frac{6}{929}\)

\(\frac{3}{2}A=\frac{2}{15}+\frac{2}{35}+\frac{2}{63}+\frac{2}{99}+\frac{2}{???}\)

Bạn nên xem lại bài 

S = 1/9 + 1/45 + 1/105 + 1/189 + 1/297

=> S = 1/2 ( 6/27 + 6/135 + 6/315 + 6/567 + 6/891 )

=> S = 1/2 ( 6/3.9 + 6/9.15 + 6/15.21 + 6/21.27 + 6/27.33 )

=> S = 1/2 ( 1/3 - 1/9 + 1/9 - 1/15 + ... + 1/27 - 1/33 )

=> S = 1/2 ( 1/3 - 1/33 )

=> S = 1/2 . 10/33

=> S = 5/33

\(S=\frac{1}{9}+\frac{1}{45}+\frac{1}{105}+\frac{1}{189}+\frac{1}{297}\)

\(S=\frac{1}{1.9}+\frac{1}{9.5}+\frac{1}{5.21}+\frac{1}{21.9}+\frac{1}{9.33}\)

\(5S=\frac{5}{1.9}+\frac{5}{9.5}+\frac{5}{5.21}+\frac{5}{21.9}+\frac{5}{9.33}\)

\(5S=1-\frac{1}{9}+\frac{1}{9}-\frac{1}{5}+\frac{1}{5}+\frac{1}{21}+\frac{1}{21}-\frac{1}{9}+\frac{1}{9}-\frac{1}{33}\)

\(5S=1-\frac{1}{33}\)

\(5S=\frac{32}{33}\)

\(S=\frac{32}{33}:5\)

\(S=\frac{32}{165}\)

Cậu Bé Tiến Pro sai rồi nhé!

\(\frac{5}{1.9}=1-\frac{1}{9}\Leftrightarrow9-1=5\)

-.-

S+..... 

là sao vậy ??

S = NHA

\(=\frac{5}{33}\)

@@@@@@@@@@@

~ học tốt ~

\(\frac{4}{9}+\frac{4}{45}+\frac{4}{105}+\frac{4}{189}\)

\(=\frac{8}{15}+\frac{4}{105}+\frac{4}{189}\)

\(=\frac{4}{7}+\frac{4}{189}\)

\(=\frac{16}{27}\)

Học tốt #

\(S=\left(1+4\right)+\left(4^2+4^3\right)+...+\left(4^{98}+4^{99}\right)\\ S=\left(1+4\right)+4^2\left(1+4\right)+...+4^{98}\left(1+4\right)\\ S=\left(1+4\right)\left(1+4^2+...+4^{98}\right)=5\left(1+4^2+...+4^{98}\right)⋮5\)

\(S=\left(1+4\right)+...+4^{98}\left(1+4\right)\)

\(=5\left(1+...+4^{98}\right)⋮5\)

\(S=\left(1+3\right)+...+3^8\left(1+3\right)\)

\(=4\left(1+...+3^8\right)⋮4\)