\(\frac{C}{9}=\frac{4}{1.3.5}+\frac{4}{3.5.7}+\frac{4}{5.7.9}+...+\frac{4}{21.23.25}+\frac{4}{23.25.27}.\)

\(\frac{C}{9}=\frac{5-1}{1.3.5}+\frac{7-3}{3.5.7}+\frac{9-5}{5.7.9}+...+\frac{25-21}{21.23.25}+\frac{27-23}{23.25.27}\)

\(\frac{C}{9}=\frac{1}{1.3}-\frac{1}{3.5}+\frac{1}{3.5}-\frac{1}{5.7}+\frac{1}{5.7}-\frac{1}{7.9}+...+\frac{1}{21.23}-\frac{1}{23.25}+\frac{1}{23.25}-\frac{1}{25.27}\)

\(\frac{C}{9}=\frac{1}{3}-\frac{1}{25.27}\Rightarrow C=\frac{9\left(25.9-1\right)}{25.27}=\frac{25.9-1}{25.3}=3-\frac{1}{25.3}< 3\)

\(\dfrac{5}{9}+\dfrac{7}{12}-\dfrac{3}{4}=\dfrac{5.4}{36}+\dfrac{7.3}{36}-\dfrac{3.9}{36}\\ =\dfrac{20+21-27}{36}\\ =\dfrac{14}{36}=\dfrac{7}{18}\)

chứng minh B làm sao z

Cách 1

\(2\dfrac{3}{4}\)+ \(5\dfrac{7}{9}\)= \(\dfrac{11}{4}+\dfrac{47}{9}\)

= \(\dfrac{99}{36}+\dfrac{188}{36}\)

= \(\dfrac{287}{36}\)= 7,9722...

Cách 2

\(2\dfrac{3}{4}+5\dfrac{7}{8}=2\dfrac{55}{36}+5\dfrac{55}{36}\)

= \(7\dfrac{55}{36}=8\dfrac{19}{36}\)

Cách 1 mình làm nhầm bài đúng phải là

\(2\dfrac{3}{4}+5\dfrac{7}{9}=\dfrac{11}{4}+\dfrac{52}{9}=\dfrac{99}{36}+\dfrac{208}{36}\)

=\(\dfrac{307}{36}\)=8,527777778

\(\frac{36}{1\cdot3\cdot5}+\frac{36}{3\cdot5\cdot7}+\frac{36}{5\cdot7\cdot9}+...+\frac{36}{25\cdot27\cdot29}\)

\(=9\left[\frac{4}{1\cdot3\cdot5}+\frac{4}{3\cdot5\cdot7}+\frac{4}{5\cdot7\cdot9}+...+\frac{4}{25\cdot27\cdot29}\right]\)

\(=9\left[\frac{1}{1\cdot3}-\frac{1}{3\cdot5}+\frac{1}{3\cdot5}-\frac{1}{5\cdot7}+...+\frac{1}{25\cdot27}-\frac{1}{27\cdot29}\right]\)

\(=9\left[\frac{1}{3}-\frac{1}{783}\right]=9\cdot\frac{260}{783}=\frac{260}{87}\)

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

\(\Rightarrow\frac{1}{9}A=\frac{4}{1.3.5}+\frac{4}{3.5.7}+\frac{4}{5.7.9}+...+\frac{4}{25.27.29}\)

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

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

\(\Rightarrow\frac{1}{9}A=\frac{261}{783}-\frac{1}{783}\)

\(\Rightarrow\frac{1}{9}A=\frac{260}{783}\)

\(\Rightarrow A=\frac{260}{783}\div\frac{1}{9}\)

\(\Rightarrow A=\frac{2340}{783}=\frac{260}{87}\)

a, A= 1/2. (2/1.2.3+2/2.3.4+2/3.4.5+...+2/18.19.20) A=1/2. (1/1.2-1/2.3+1/2.3-1/3.4+1/3.4-1/4.5+...+1/18.19-1/19.20) A=1/2. (1/1.2-1/19.20) A=1/2. 189/380 A= 189/760

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

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

\(\Rightarrow B<3\)

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

\(=9\left(\frac{4}{3.\left(1.5\right)}+\frac{4}{5\left(3.7\right)}+\frac{4}{7.\left(5.9\right)}+.....+\frac{4}{27\left(25.29\right)}\right)\)

\(=9\left(\frac{1}{3}\left(1-\frac{1}{5}\right)+\frac{1}{5}\left(\frac{1}{3}-\frac{1}{7}\right)+\frac{1}{7}\left(\frac{1}{5}-\frac{1}{9}\right)+...+\frac{1}{27}\left(\frac{1}{25}-\frac{1}{29}\right)\right)\)

\(=9\left(\frac{1}{3}-\frac{1}{3.5}+\frac{1}{5.3}-\frac{1}{5.7}+\frac{1}{7.5}-\frac{1}{7.9}+...+\frac{1}{27.25}-\frac{1}{27.29}\right)\)

\(=9\left(\frac{1}{3}-\frac{1}{27.29}\right)< 9.\frac{1}{3}=3\)

Vậy A < 3.