ai đó kết bạn với mình nha mình hết lời rùi

Áp dụng: \(\frac{4}{n\left(n+2\right)\left(n+4\right)}=\frac{n+4-n}{n\left(n+2\right)\left(n+4\right)}=\frac{1}{n\left(n+2\right)}-\frac{1}{\left(n+2\right)\left(n+4\right)}\)

\(\frac{B}{9}=\frac{1}{1.3}-\frac{1}{3.5}+\frac{1}{3.5}-\frac{1}{5.7}+...+\frac{1}{25.27}-\frac{1}{27.29}=\frac{1}{3}-\frac{1}{27.29}

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

chứng minh B làm sao z

Ta có :

\(B=\dfrac{36}{1.3.5}+\dfrac{36}{3.5.7}+\dfrac{36}{5.7.9}+...............+\dfrac{36}{25.27.29}\)

\(B=9\left(\dfrac{4}{1.3.5}+\dfrac{4}{3.5.7}+\dfrac{4}{5.7.9}+.............+\dfrac{4}{25.27.29}\right)\)

\(B=9\left(\dfrac{1}{1.3}-\dfrac{1}{3.5}+\dfrac{1}{3.5}-\dfrac{1}{5.7}+\dfrac{1}{5.7}-\dfrac{1}{7.9}+...........+\dfrac{1}{25.27}-\dfrac{1}{27.29}\right)\)

\(B=9\left(\dfrac{1}{1.3}-\dfrac{1}{27.29}\right)\)

\(B=9\left(\dfrac{1}{3}-\dfrac{1}{783}\right)\)

\(B=9.\dfrac{1}{3}-9.\dfrac{1}{783}\)

\(B=3-\dfrac{9}{783}< 3\)

\(\Rightarrow B< 3\rightarrowđpcm\)

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

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

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

\(A=9.\frac{260}{87}=\frac{260}{87}