Thao quy ước của 1 phân số lớn hơn 0 thì:

\(\dfrac{a}{b}>0=>\dfrac{a}{b}< \dfrac{a+n}{b+n}\left(n\ne0\right)\)

Áp dụng vào từng phân số trên ta có: ( các phân số trên lớn hơn 0 nên):

để ý rằng các phân số trên đều lớn hơn 1/100

=>tích cũng lớn hơn 1/100

=>A>1/100

CHÚC BẠN HỌC TỐT.............

ta có :
1/2 < 2/3
2/3 <3/4
.........
9999/10000 < 10000/10001
suy ra : A2 < 1/22/33/4*****9999/1000010000/10001
suy ra : A2 < 1/10001 < 1/10000= (1/100)2
suy ra A2 < (1/100)2 . Từ đó: A < 1/100

2 là mũ 2 nha bạn

Ta có: \(C=\dfrac{1}{2}\cdot\dfrac{3}{4}\cdot\dfrac{5}{6}...\dfrac{9999}{10000}\)

\(C=\dfrac{1\cdot3\cdot5...9999}{2\cdot4\cdot6...10000}\)

Gọi \(D=\dfrac{2}{3}\cdot\dfrac{4}{5}\cdot\dfrac{6}{7}...\dfrac{10000}{10001}\)

Mà \(\dfrac{1}{2}< \dfrac{2}{3};\dfrac{3}{4}< \dfrac{4}{5};\dfrac{5}{6}< \dfrac{6}{7};...;\dfrac{9999}{10000}< \dfrac{10000}{10001}\)

\(\Rightarrow\dfrac{1}{2}\cdot\dfrac{3}{4}\cdot\dfrac{5}{6}...\dfrac{9999}{10000}< \dfrac{2}{3}\cdot\dfrac{4}{5}\cdot\dfrac{6}{7}...\dfrac{10000}{10001}\)

\(\Rightarrow C< D\)

Ta lại có: \(C\cdot D=\left(\dfrac{1}{2}\cdot\dfrac{3}{4}\cdot\dfrac{5}{6}...\dfrac{9999}{10000}\right)\left(\dfrac{2}{3}\cdot\dfrac{4}{5}\cdot\dfrac{6}{7}...\dfrac{10000}{10001}\right)\)

\(\Rightarrow C\cdot D=\dfrac{1}{2}\cdot\dfrac{2}{3}\cdot\dfrac{3}{4}\cdot\dfrac{4}{5}\cdot\dfrac{5}{6}\cdot\dfrac{6}{7}...\dfrac{9999}{10000}\cdot\dfrac{10000}{10001}\)

\(\Rightarrow C\cdot D=\dfrac{1\cdot2\cdot3\cdot4\cdot5\cdot6...9999\cdot10000}{2\cdot3\cdot4\cdot5\cdot6\cdot7...10000\cdot10001}\)

\(\Rightarrow C\cdot D=\dfrac{1}{10001}\)

Mà \(C< D\)

\(\Rightarrow C\cdot C< C\cdot D\)

\(\Rightarrow C\cdot C< \dfrac{1}{10001}\)

\(\Rightarrow C< \dfrac{1}{10001}\)

Mà \(\dfrac{1}{10001}< \dfrac{1}{100}\)

\(\Rightarrow C< \dfrac{1}{100}\)

Vậy \(C< \dfrac{1}{100}\)

C = \(\dfrac{1}{2}\).\(\dfrac{3}{4}\).\(\dfrac{5}{6}\)....\(\dfrac{9999}{10000}\)

C < \(\dfrac{2}{3}.\dfrac{4}{5}.\dfrac{6}{7}....\dfrac{10000}{10001}\)

C² < \(\dfrac{1.\left(3.5.7...9999\right)}{\left(2.4.6...10000\right)}.\dfrac{\left(2.4.6...10000\right)}{\left(3.5.7...9999\right).10001}\)

C^{2 < \(\dfrac{1}{10001}\)}

C² < \(\left(\dfrac{1}{100}\right)^2\)

C < \(\dfrac{1}{100}\)

Vậy \(\dfrac{1}{2}.\dfrac{3}{4}.\dfrac{5}{6}....\dfrac{9999}{10000}< \dfrac{1}{100}\)

Chúc bạn học tốt !

ta có : 1/2 < 2/3

2/3 <3/4

......... 9999/10000 < 10000/10001 suy ra : A2 < 1/22/33/4*****9999/1000010000/10001 suy ra : A2 < 1/10001 < 1/10000= (1/100)2 suy ra A2 < (1/100)2 . Từ đó: A < 1/100 = 0,01

a: \(\dfrac{4^5\cdot9^4-2\cdot6^9}{2^{10}\cdot3^8+6^8\cdot20}\)

\(=\dfrac{2^{10}\cdot3^8-2\cdot2^9\cdot3^9}{2^{10}\cdot3^8+2^8\cdot3^8\cdot2^2\cdot5}\)

\(=\dfrac{2^{10}\cdot3^8-2^{10}\cdot3^9}{2^{10}\cdot3^8+2^{10}\cdot3^8\cdot5}\)

\(=\dfrac{2^{10}\cdot3^8\left(1-3\right)}{2^{10}\cdot3^8\left(1+5\right)}=\dfrac{-2}{6}=-\dfrac{1}{3}\)

Giải:

Vì:

\(\dfrac{1}{2}< \dfrac{2}{3}.\)

\(\dfrac{2}{3}< \dfrac{3}{4}.\)

.............

\(\dfrac{9999}{10000}< \dfrac{10000}{10001}.\)

\(\Rightarrow S^2< \dfrac{1}{2}.\dfrac{2}{3}.\dfrac{3}{4}.....\dfrac{9999}{10000}.\dfrac{10000}{10001}.\)

\(\Rightarrow S^2< \dfrac{1}{10001}< \dfrac{1}{10000}=\left(\dfrac{1}{100}\right)^2.\)

Mà \(\left(\dfrac{1}{100}\right)^2=0,01^2.\)

\(\Rightarrow S^2< 0,01^2\left(=\left(\dfrac{1}{100}\right)^2\right).\)

\(\Rightarrow S< 0,01.\)

Vậy \(S< 0,01.\)

~ Học tốt!!! ~