\(A=9\left(\dfrac{4}{1\cdot3\cdot5}+\dfrac{4}{3\cdot5\cdot7}+...+\dfrac{4}{25\cdot27\cdot29}\right)\)

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

\(=9\left(\dfrac{1}{3}-\dfrac{1}{783}\right)=\dfrac{260}{87}\)

ai đó kết bạn với mình nha mình hết lời rùi

Gọi B là biểu thức đã cho.

ta chứng minh:

\(\frac{4}{1.3.5}=\frac{1}{1.3}-\frac{1}{3.5}\)

\(\frac{4}{3.5.7}=\frac{1}{3.5}-\frac{1}{5.7}\)

....

Ta có:

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

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

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

\(=3-\frac{1}{27.29}<3\) (điều phải cm)

Gọi B là biểu thức đã cho.
Dễ dàng chứng minh:

...
Ta có:
B

Ta có :

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

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

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

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

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

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

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

\(\Rightarrow B< 3\rightarrowđpcm\)

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

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

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

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