https://lazi.vn/edu/exercise/289338/cho-b-2-1-x-4-3-x-6-5-x-8-x-x-200-199-chung-minh-rang-14-lt-b-lt-20

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Ta có

A\(^2\) = \(\left(\frac{2}{1}.\frac{4}{3}.....\frac{200}{199}\right)^2=\left(\frac{2}{1}.\frac{4}{3}.....\frac{200}{199}\right).\left(\frac{2}{1}.\frac{4}{3}.....\frac{200}{199}\right)\)

\(\Rightarrow A^2 >\left(\frac{2}{1}.\frac{4}{3}...\frac{200}{199}\right).\left(\frac{3}{2}.\frac{5}{4}...\frac{201}{200}\right)\)

( Áp dụng tính chất \(\frac{n+1}{n}>\frac{n+3}{n+1}\forall\)n dương )

\(\Rightarrow A^2>201>196=14^2\)

\(\Rightarrow A>14\left(2\right)\)

Lại có :

\(A^2=\left(\frac{2}{1}.\frac{4}{3}....\frac{200}{199}\right).\left(\frac{2}{1}.\frac{4}{3}....\frac{200}{199}\right)\)

\(\Rightarrow A^2< 2.\left(\frac{2}{1}.\frac{4}{3}...\frac{200}{199}\right).\left(\frac{3}{2}.\frac{5}{4}...\frac{199}{198}\right)\)

\(\Rightarrow A^2< 2.200=400=20^2\)

\(\Rightarrow A< 20\left(2\right)\)

Từ (1) và (2) => 14 < A < 20

Bài mình còn sai sót đôi chỗ ko biết bạn có hiểu ko ạ

No choice teenAkai Harumacô giải hộ e vs ak

thnk

\(A=\frac{1}{199}+\frac{2}{198}+\frac{3}{197}+...+\frac{189}{2}+\frac{199}{1}\)

\(A=\frac{1}{199}+\frac{2}{198}+\frac{3}{197}+...+\frac{198}{2}+199\)

\(A=\left(\frac{1}{199}+1\right)+\left(\frac{2}{198}+1\right)+\left(\frac{3}{197}+1\right)+...+\left(\frac{198}{2}+1\right)+1\)

\(A=\frac{200}{199}+\frac{200}{198}+\frac{200}{197}+...+\frac{200}{2}+1\)

\(A=\frac{200}{200}+\frac{200}{199}+\frac{200}{198}+\frac{200}{197}+...+\frac{200}{2}\)

\(A=200\left(\frac{1}{200}+\frac{1}{199}+\frac{1}{198}+...+\frac{1}{2}\right)\)

Vậy    \(A=200\left(\frac{1}{200}+\frac{1}{199}+\frac{1}{198}+...+\frac{1}{2}\right)\)