\(\dfrac{3}{2.6}\) + \(\dfrac{3}{6.10}\) + \(\dfrac{3}{10.14}\)

=  \(\dfrac{3}{4}\).(\(\dfrac{4}{2.6}\) + \(\dfrac{4}{6.10}\) + \(\dfrac{4}{10.14}\))

= \(\dfrac{3}{4}\).(\(\dfrac{1}{2}-\dfrac{1}{6}\) + \(\dfrac{1}{6}\) - \(\dfrac{1}{10}\) + \(\dfrac{1}{10}\) - \(\dfrac{1}{14}\))

= \(\dfrac{3}{4}\).(\(\dfrac{1}{2}\) - \(\dfrac{1}{14}\))

= \(\dfrac{3}{4}\). \(\dfrac{3}{7}\)

= \(\dfrac{9}{28}\)

B = \(\dfrac{4}{1.3.5}\) + \(\dfrac{4}{3.5.7}\) + \(\dfrac{4}{5.7.9}\)

B = \(\dfrac{1}{1.3}\) - \(\dfrac{1}{3.5}\) + \(\dfrac{1}{3.5}\) - \(\dfrac{1}{5.7}\) + \(\dfrac{1}{5.7}\) - \(\dfrac{1}{7.9}\)

B = \(\dfrac{1}{1.3}\) - \(\dfrac{1}{7.9}\)

B = \(\dfrac{1}{3}\) - \(\dfrac{1}{63}\)

B =  \(\dfrac{20}{63}\)

C = \(\dfrac{5}{2.4.6}\) + \(\dfrac{5}{4.6.8}\) + \(\dfrac{5}{6.8.10}\)

C = \(\dfrac{5}{4}\).(\(\dfrac{4}{2.4.6}\) + \(\dfrac{4}{4.6.8}\) + \(\dfrac{4}{6.8.10}\))

C = \(\dfrac{5}{4}\).(\(\dfrac{1}{2.4}\) - \(\dfrac{1}{4.6}\) + \(\dfrac{1}{4.6}\) - \(\dfrac{1}{6.8}\) + \(\dfrac{1}{6.8}\) - \(\dfrac{1}{8.10}\))

C = \(\dfrac{5}{4}\).(\(\dfrac{1}{2.4}\) - \(\dfrac{1}{8.10}\))

C = \(\dfrac{5}{4}\).( \(\dfrac{1}{8}\) - \(\dfrac{1}{80}\))

C = \(\dfrac{5}{4}\). \(\dfrac{9}{80}\)

C = \(\dfrac{9}{64}\)

\(A=\frac{1}{1.3}-\frac{1}{3.5}+\frac{1}{3.5}-\frac{1}{5.7}+...+\frac{1}{1997.1999}-\frac{1}{1999.2001}\)

     \(=\frac{1}{1.3}-\frac{1}{1999.2001}\)

       Bạn tính kết quả nhé

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!

chứng minh B làm sao z

$\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)}$4n(n+2)(n+4) =n+4−nn(n+2)(n+4) =1n(n+2) −1(n+2)(n+4) $\frac{B}{9}=\frac{1}{1.3}-\frac{1}{3.5}+\frac{1}{3.5}-\frac{1}{5.7}+...+\frac{1}{25.27}-\frac{1}{27.29}=\frac{1}{3}-\frac{1}{27.29}<\frac{1}{3}$B9 =11.3 −13.5 +13.5 −15.7 +...+125.27 −127.29 =13 −127.29 <13 $\Rightarrow B<3$

=1/1.3-1/3.5+1/3.5-1/5.7+...+1/99.11-1/11.13

=1/1.3-1/11.13

=1/3-1/143

=140/429

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}