Ta có \(A=\frac{1}{3}.\frac{4}{6}.\frac{7}{9}.....\frac{208}{210}=\frac{1.4.7.....208}{3.6.9.....210}\)

Mà \(1< 3,4< 6,7< 9,...,208< 210\)

\(\Rightarrow1.4.7.....208< 3.6.9.....210\)

\(\Rightarrow\frac{1.4.7.....208}{3.6.9.....210}< 1\)\(\Leftrightarrow A< 1\)

Lại có \(1< 25\)\(\Rightarrow A< 25\)

A= 1/1.2 + 1/2.3 + 1/3.4 + 1/4.5 + 1/5.6 + 1/6.7 + 1/ 7.8 + 1/ 8.9 + 1/ 9.10 

=> A = 1/1 - 1/2 + 1/2 - 1/3 + 1/3 - 1/4 + 1/4 - 1/5 + 1/5 - 1/6 + 1/6 - 1/7 + 1/7 - 1/8 + 1/8 - 1/9 + 1/9 - 1/10

=> A = 1 - 1/10 = 9/10

Vậy A = 9/10

A = 1 - 1/2 + 1/2 - 1/3 + 1/3 - 1/4 + 1/4 - 1/5 + 1/5 - 1/6 + 1/6 - 1/7 + 1/7 - 1/8 + 1/8 - 1/9 + 1/9 - 1/10

A = 1 - 1/10 = 9/10

A= 1/1.2 + 1/2.3 + 1/3.4 + 1/4.5 + 1/5.6 + 1/6.7 + 1/ 7.8 + 1/ 8.9 + 1/ 9.10

= 1/1 - 1/2 + 1/2 - 1/3 + 1/3 - 1/4 + 1/4 - 1/5 + 1/5 - 1/6 + 1/6 - 1/7 + 1/7 - 1/8 + 1/8 - 1/9 + 1/9 - 1/10

= 1 - 1/10 = 9/10

Công thức viết khó đọc quá. Bạn nên viết đề bằng công thức toán (biểu tượng $\sum$ góc trái khung soạn thảo để được hỗ trợ tốt hơn.

B = \(\dfrac{1}{3.4}\) - \(\dfrac{1}{4.5}\) - \(\dfrac{1}{5.6}\) - \(\dfrac{1}{6.7}\) - \(\dfrac{1}{7.8}\) - \(\dfrac{1}{8.9}\) - \(\dfrac{1}{9.10}\)

B = \(\dfrac{1}{12}\) - ( \(\dfrac{1}{4.5}\) + \(\dfrac{1}{5.6}\) + \(\dfrac{1}{6.7}\) + \(\dfrac{1}{7.8}\) + \(\dfrac{1}{8.9}\) + \(\dfrac{1}{9.10}\))

B = \(\dfrac{1}{12}\) - ( \(\dfrac{1}{4}\) - \(\dfrac{1}{5}\) + \(\dfrac{1}{5}\) - \(\dfrac{1}{6}\) + \(\dfrac{1}{6}\) - \(\dfrac{1}{7}\) + \(\dfrac{1}{7}\) - \(\dfrac{1}{8}\) + \(\dfrac{1}{8}\) - \(\dfrac{1}{9}\) + \(\dfrac{1}{9}\) - \(\dfrac{1}{10}\))

B =  \(\dfrac{1}{12}\) - ( \(\dfrac{1}{4}\) - \(\dfrac{1}{10}\)) 

B = \(\dfrac{1}{12}\) - \(\dfrac{3}{20}\) 

B = - \(\dfrac{1}{15}\)

Có:

\(\dfrac{n}{n+2}< \dfrac{n-1}{n}\)(Vì
\(n^2< n^2+n-2\forall n>2\))

Nên ta có

\(F=\dfrac{1}{3}.\dfrac{4}{6}....\dfrac{208}{201}\)

\(\Rightarrow F< \dfrac{1}{3}.\dfrac{3}{4}.\dfrac{6}{7}...\dfrac{207}{208}\)

\(\Rightarrow F^2< \dfrac{1.4.7...208}{3.6.9.12...210}.\dfrac{1.3.6.9...207}{3.4.7.10.208}\)

\(\Rightarrow F^2=\dfrac{1}{210}.\dfrac{1}{3}\)

\(\Rightarrow F^2=\dfrac{1}{630}< \left(\dfrac{1}{25}\right)^2\)

Vậy F\(< \dfrac{1}{25}\)