\(A=\frac{24}{1.3.5}+\frac{24}{3.5.7}+\frac{24}{5.7.9}+...+\frac{24}{25.27.29}\)

\(A=6\left(\frac{4}{1.3.5}+\frac{4}{3.5.7}+\frac{4}{5.7.9}+...+\frac{4}{25.27.29}\right)\)

\(A=6\left(\frac{5-1}{1.3.5}+\frac{7-3}{3.5.7}+\frac{9-5}{5.7.9}+...+\frac{29-25}{25.27.29}\right)\)

\(A=6\left(\frac{1}{1.3}-\frac{1}{3.5}+\frac{1}{3.5}-\frac{1}{5.7}+...+\frac{1}{25.27}-\frac{1}{27.29}\right)\)

\(A=6\left(\frac{1}{1.3}-\frac{1}{27.29}\right)=\frac{520}{261}\)

ta có 

\(A=6\left(\frac{4}{1.3.5}+\frac{4}{3.5.7}+..+\frac{4}{25.27.29}\right)=6\left(\frac{5-1}{1.3.5}+\frac{7-3}{3.5.7}+..+\frac{29-25}{25.27.29}\right)\)

\(=6\left(\frac{1}{1.3}-\frac{1}{3.5}+\frac{1}{3.5}-\frac{1}{5.7}+..+\frac{1}{25.27}-\frac{1}{27.29}\right)=6\left(\frac{1}{3}-\frac{1}{27.29}\right)\)

\(=2-\frac{2}{9.29}=\frac{520}{261}\)

Đặt tổng là A

\(\frac{A}{6}=\frac{4}{1.3.5}+\frac{4}{3.5.7}+\frac{4}{5.7.9}+...+\frac{6}{25.27.29}\)

\(\frac{A}{6}=\frac{5-1}{1.3.5}+\frac{7-3}{3.5.7}+\frac{9-5}{5.7.9}+...+\frac{29-25}{25.27.29}\)

\(\frac{A}{6}=\frac{1}{1.3}-\frac{1}{3.5}+\frac{1}{3.5}-\frac{1}{5.7}+\frac{1}{5.7}-\frac{1}{7.9}+...+\frac{1}{25.27}-\frac{1}{27.29}\)

\(\frac{A}{6}=\frac{1}{1.3}-\frac{1}{27.29}\Rightarrow A=\left(\frac{1}{3}-\frac{1}{27.29}\right):6\)

B =  9 . [ 4/1.3.5+4/3.5.7+4/5.7.9+...+4/25.27.29]

B =  9 . [ 1/3-1/783]

   =  9 . [ 1/3-1/783] 

   =  9 . 260/783=260/87<261/87<3

Bạn sai rồi 261/87=3 mà

ai đó kết bạn với mình nha mình hết lời rùi

Ta có:

\(A=\frac{36}{1.3.5}+\frac{36}{3.5.7}+\frac{36}{5.7.9}+...+\frac{36}{25.27.29}\)

\(\Rightarrow A=9.\left(\frac{4}{1.3.5}+\frac{4}{3.5.7}+\frac{4}{5.7.9}+...+\frac{4}{25.27.29}\right)\)

\(\Rightarrow A=9.\left(\frac{1}{1.3}-\frac{1}{3.5}+\frac{1}{3.5}-\frac{1}{5.7}+\frac{1}{5.7}-\frac{1}{7.9}+...+\frac{1}{25.27}-\frac{1}{27.29}\right)\)

\(\Rightarrow A=9.\left(\frac{1}{1.3}-\frac{1}{27.29}\right)\)

\(\Rightarrow A=9.\left(\frac{1}{3}-\frac{1}{783}\right)\)

\(\Rightarrow A=9.\frac{1}{3}-9.\frac{1}{783}\)

\(\Rightarrow A=3-\frac{1}{87}\)

Vì \(3-\frac{1}{87}< 3.\)

\(\Rightarrow A< 3\left(đpcm\right).\)

Chúc bạn học tốt!