Xét với mọi n > 2 , ta có \(\frac{n}{n+2}< \frac{n-1}{n}\) (vì \(n^2< n^2+n-2\))

Áp dụng : \(A=\frac{1}{3}.\frac{4}{6}.\frac{7}{9}.\frac{10}{12}...\frac{208}{210}< \frac{1}{3}.\frac{3}{4}.\frac{6}{7}.\frac{9}{10}...\frac{207}{208}\)

Suy ra : \(A^2< \frac{1.4.7.10...208}{3.6.9.12...210}.\frac{1.3.6.9...207}{3.4.7.10...208}=\frac{1}{210}.\frac{1}{3}=\frac{1}{630}< \frac{1}{625}=\left(\frac{1}{25}\right)^2\)

Do đó \(A< \frac{1}{25}\)

hiểu j chết liền

=="

đúng tick cho

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\(< \dfrac{1}{25}\)