Đặt A = \(\frac{2}{3}.\frac{4}{5}.\frac{6}{7}...\frac{9998}{9999}.\frac{10000}{10000}\)

Rõ ràng A < A'

=> A² < A . A' \(=\frac{1}{10000}=\frac{1}{100^2}\)

Nên A < 0,01

A<2/3*4/5*6/7...10000/10001

A^2<A*(2/3*4/5*6/7...10000/10001)

A^2<\(\frac{1\cdot2\cdot3\cdot4\cdot5\cdot6...9999\cdot10000}{2\cdot3\cdot4\cdot5\cdot6\cdot7...10000\cdot10001}\)

A^2<1/10001

0,01=1/100

1/100^2=1/10000

A^2<1/10001<1/10000

lon hon la cai chac

A lon hon B

So sánh:

Ta có:

A > 0,01

Ta có: \(0,01=\frac{1}{100}\)

Mà \(A=\frac{1}{2}.\frac{3}{4}.\frac{5}{6}.....\frac{9999}{10000}\)

Ta thấy: \(\frac{1}{100}=\frac{100}{10000}\)

Vì \(\frac{9999}{10000}>\frac{100}{10000}hay\frac{9999}{10000}>\frac{1}{100}\)

Nên \(\frac{1}{2}.\frac{3}{4}.\frac{5}{6}.....\frac{9999}{10000}>\frac{1}{100}hay\frac{1}{2}.\frac{3}{4}.\frac{5}{6}.....\frac{9999}{10000}>0,01\)

Vậy \(A>0,01\)

Ta có: \(0,01=\frac{1}{100}\)

Đặt  \(B=\frac{2}{3}.\frac{4}{5}.\frac{6}{7}....\frac{10000}{10001}\)

Xét  \(AB=\frac{1}{2}.\frac{3}{4}.\frac{5}{6}....\frac{9999}{10000}.\frac{2}{3}.\frac{4}{5}.\frac{6}{7}....\frac{10000}{10001}\)

\(\Leftrightarrow\)\(AB=\frac{1.2.3.4.5.6.....9999.10000}{2.3.4.5.6.7.....10000.10001}\)

\(\Leftrightarrow\)\(AB=\frac{1}{10001}\)

​Vì A < B

\(\Rightarrow\)A² < AB

\(\Rightarrow A^2< \frac{1}{10001}< \frac{1}{10000}\)

\(\Rightarrow A< \frac{1}{100}hayA< 0,01\)

Vậy A < 0,01

https://h.vn/hoi-dap/question/203861.html

bạn tham khảo nhé

Có A = 1/2.3/4.5/6 ... 9999/10000 

Đặt B = 2/3.4/5.6/7 ... 10000/10001
Ta có A.B = 1/2.2/3.3/4 ... 10000/10001 = 1/10001 (1)
Lại có :
1/2 < 2/3
3/4 < 4/5
................
9999/10000 < 10000/10001

 =>1/2.3/4.....9999/10000<2/3.4/5.....10000/10001

=> A < B => A² < A.B (2)
(1),(2) => A² < 1/10001 => A²<1/10000=>A<1/100=0,01(đpcm)

#Chino

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