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A = \(\frac{2.4.6+4.6.8+6.8.10+8.10.12+...+198.200.202}{1.3.5+3.5.7+5.7.9+7.9.11+...+97.99.101}\) =?
\(A=\frac{2}{2.4.6}+\frac{2}{4.6.8}+\frac{2}{6.8.10}+\frac{2}{8.10.12}\)
\(A=\frac{2}{48}+\frac{2}{192}+\frac{2}{480}+\frac{2}{960}\)
\(A=\frac{1}{24}+\frac{1}{96}+\frac{1}{240}+\frac{1}{480}\)
\(A=\frac{20}{480}+\frac{5}{480}+\frac{2}{480}+\frac{1}{480}\)
\(A=\frac{7}{120}\)
A = \(\dfrac{2}{2.4.6}\) + \(\dfrac{2}{4.6.8}\) + \(\dfrac{2}{6.8.10}\) + \(\dfrac{2}{8.10.12}\)
A = \(\dfrac{2}{2}\).(\(\dfrac{2}{2.4.6}\) + \(\dfrac{2}{4.6.8}\) + \(\dfrac{2}{6.8.10}\) + \(\dfrac{2}{8.10.12}\))
A = \(\dfrac{1}{2}\).(\(\dfrac{2.2}{2.4.6}\) + \(\dfrac{2.2}{4.6.8}\) + \(\dfrac{2.2}{6.8.10}+\dfrac{2.2}{8.10.12}\))
A = \(\dfrac{1}{2}\).( \(\dfrac{4}{2.4.6}+\dfrac{4}{4.6.8}+\dfrac{4}{6.8.10}+\dfrac{4}{8.10.12}\))
A = \(\dfrac{1}{2}\).(\(\dfrac{1}{2.4}\) - \(\dfrac{1}{4.6}\) +\(\dfrac{1}{4.6}\) - \(\dfrac{1}{6.8}\) + \(\dfrac{1}{6.8}\) - \(\dfrac{1}{8.10}\) + \(\dfrac{1}{8.10}\) - \(\dfrac{1}{10.12}\))
A = \(\dfrac{1}{2}\).(\(\dfrac{1}{2.4}\) - \(\dfrac{1}{10.12}\))
A = \(\dfrac{1}{2}\).(\(\dfrac{1}{8}-\dfrac{1}{120}\))
A = \(\dfrac{1}{2}\).\(\dfrac{7}{60}\)
A = \(\dfrac{7}{120}\)
$\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 $\Rightarrow B<3$
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}
a)\(A=\frac{36}{1.3.5}+\frac{36}{3.5.7}+\frac{36}{5.7.9}+...+\frac{36}{25.27.29}\)
=\(\frac{9.4}{1.3.5}+\frac{9.4}{3.5.7}+\frac{9.4}{5.7.9}+...+\frac{9.4}{25.27.29}\)
=\(9.\left(\frac{4}{1.3.5}+\frac{4}{3.5.7}+\frac{4}{5.7.9}+...+\frac{4}{25.27.29}\right)\)
=\(9.\left(\frac{1}{1.3}-\frac{1}{3.5}+\frac{1}{3.5}-\frac{1}{5.7}+\frac{1}{5.7}-\frac{1}{7.9}+...+\frac{1}{25.27}-\frac{1}{27.29}\right)\)
=\(9.\left(\frac{1}{3}-\frac{1}{27.29}\right)=9.\left(\frac{1}{3}-\frac{1}{783}\right)=9.\left(\frac{261}{783}-\frac{1}{783}\right)=9.\frac{260}{783}\)
=\(\frac{260}{87}\)
b)
ta có: \(3=\frac{261}{87}>\frac{260}{87}\)
vậy A<3
\(\dfrac{3}{2.6}\) + \(\dfrac{3}{6.10}\) + \(\dfrac{3}{10.14}\)
= \(\dfrac{3}{4}\).(\(\dfrac{4}{2.6}\) + \(\dfrac{4}{6.10}\) + \(\dfrac{4}{10.14}\))
= \(\dfrac{3}{4}\).(\(\dfrac{1}{2}-\dfrac{1}{6}\) + \(\dfrac{1}{6}\) - \(\dfrac{1}{10}\) + \(\dfrac{1}{10}\) - \(\dfrac{1}{14}\))
= \(\dfrac{3}{4}\).(\(\dfrac{1}{2}\) - \(\dfrac{1}{14}\))
= \(\dfrac{3}{4}\). \(\dfrac{3}{7}\)
= \(\dfrac{9}{28}\)
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 = \(\dfrac{20}{63}\)