Đặt :
\(A=\frac{1}{2}.\frac{3}{4}.\frac{5}{6}....\frac{9999}{10000}\)
Đặt :
B=\(\frac{2}{3}.\frac{4}{5}.\frac{6}{7}.....\frac{9998}{9999}.\frac{10000}{10000}\)
Ta thấy " A<B
\(\Rightarrow A.A< A.B=\frac{1}{100^2}\\ \Rightarrow A^2< \frac{1}{100^2}\\ \Rightarrow A< \frac{1}{100}\)
Đặt \(A=\frac{1}{2}.\frac{3}{4}.\frac{5}{6}....\frac{9999}{10000}\)\(\left(A>0\right)\)
.Và \(B=\frac{2}{3}.\frac{4}{5}.\frac{6}{7}...\frac{10000}{10001}\)\(\left(B>0\right)\)
Mặt khác :
\(\frac{1}{2}< \frac{2}{3}\)
\(\frac{3}{4}< \frac{4}{5}\)
... ... ...
\(\frac{9999}{10000}< \frac{10000}{10001}\)
Nhân tất cả vế theo vế \(\Rightarrow A< B\Rightarrow A^2< A.B\left(2\right)\)
(1),(2) \(\Rightarrow A^2< \frac{1}{10001}\Rightarrow A< \sqrt{\left(\frac{1}{10001}\right)}< \sqrt{\left(\frac{1}{10000}\right)}=\frac{1}{100}\left(ĐPCM\right)\)
Ta có 27^5=3^3^5=3^15
243^3=3^5^3=3^15
Vậy A=B
2^300=2^(3.100)=2^3^100=8^100
3^200=3^(2.100)=3^2^100=9^100
Vậy A<B