Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
\(2005A=\frac{2005\left(2005^{2005}+1\right)}{2005^{2006}+1}=\frac{2005^{2006}+2005}{2005^{2006}+1}=\frac{2005^{2006}+1+2004}{2005^{2006}+1}=\frac{2005^{2006}+1}{2005^{2006}+1}+\frac{2004}{2005^{2006}+1}=1+\frac{2004}{2005^{2006}+1}\)
\(2005B=\frac{2005\left(2005^{2004}+1\right)}{2005^{2005}+1}=\frac{2005^{2005}+2005}{2005^{2005}+1}=\frac{2005^{2005}+1+2004}{2005^{2005}+1}=\frac{2005^{2005}+1}{2005^{2005}+1}+\frac{2004}{2005^{2005}+1}=1+\frac{2004}{2005^{2005}+1}\)
vì 20052006+1>20052005+1
\(\Rightarrow\frac{4}{2005^{2006}+1}< \frac{4}{2005^{2005}+1}\)
\(\Rightarrow1+\frac{4}{2005^{2006}+1}< 1+\frac{4}{2005^{2005}+1}\)
=>A<B
Ta có : A=2005^2005+1/2005^2006+1
=>2005A=2005.(2005^2005+1)/2005^2006+1
=>2005A=2005^2006+2005/2005^2006+1
=>2005A=2005^2006+1+2004/2005^2006+1
=>2005A=2005^2006+1/2005^2006+1 + 1/2005^2006+1
=>2005A=1+1/2005^2006+1
Lại có:B=2005^2004+1/2005^2005+1
=>2005B=2005.(2005^2004+1)/2005^2005+1
=>2005B=2005^2005+2005/2005^2005+1
=>2005B=2005^2005+1+2004/2005^2005+1
=>2005B=2005^2005+1/2005^2005+1 + 1/2005^2005+1
=>2005B=1+1/2005^2005+1
Vì 2006>2005
=>2005^2006>2005^2005
=>2005^2006+1>2005^2005+1
=>1/2005^2006+1<1/2005^2005+1
=>1+1/2005^2006+1<1+1/2005^2005+1
=>2005A<2005B
=>A<B
Vậy A<B
Ủng hộ mik nha mọi người !!!
\(A=\frac{2005^{2005}+1}{2005^{2006}+1}\)
\(2005A=\frac{2005^{2006}+2005}{2005^{2006}+1}=\frac{2005^{2006}+1+2004}{2005^{2006}+1}=\frac{2005^{2006}+1}{2005^{2006}+1}+\frac{2004}{2005^{2006}+1}\)
\(B=\frac{2005^{2004}+1}{2005^{2005}+1}\)
\(2005B=\frac{2005^{2005}+2005}{2005^{2005}+1}=\frac{2005^{2005}+1+2004}{2005^{2005}+1}=\frac{2005^{2005}+1}{2005^{2005}+1}+\frac{2004}{2005^{2005}+1}\)
Vì \(\frac{2004}{2005^{2006}+1}<\frac{2004}{2005^{2005}+1}\)
Nên A<B
Xét A trước ta có
\(A=\frac{2005^{2005}+1}{2005^{2006}+1}\)ta có \(2005.A=\frac{2005.\left(2005^{2005}+1\right)}{2005^{2006}+1}\)
\(2005A=\frac{2005^{2006}+2005}{2005^{2006}+1}\)\(2005A=\frac{2005^{2006}+1+2004}{2005^{2006}+1}\)
\(2005.A=1+\frac{2004}{2005^{2006}+1}\)
Xét B ta có
\(B=\frac{2005^{2004}+1}{2005^{2005}+1}\)ta có \(2005B=\frac{2005\left(2005^{2004}+1\right)}{2005^{2005}+1}\)
\(2005B=\frac{2005^{2005}+2005}{2005^{2005}+1}\)\(2005B=\frac{2005^{2005}+1+2004}{2005^{2005}+1}\)
\(2005B=1+\frac{2004}{2005^{2005}+1}\)
ta có vì 2005A<2005B
từ đó suy ra A<B
nhớ **** đó
Bài giải
A = 2005^2005 +1/ 2005^2006 + 1
suy ra ta có : 2005A = 2005^2006 + 2005 / 2005^2006 +1 = 1 +2004 / 2005^2006 + 1
B = 2005 ^ 2004 +1 / 2005 ^ 2005 +1
suy ra ta có : 2005B = 2005^2005 + 2005 / 2005^2005 +1 =1 + 2004 / 2005 ^2005 + 1
Vì 2004/2005^2006 +1 < 2004/ 2005^2005 + 1 suy ra 2005A < 2005B nên A < B
vậy A <B
\(A=\frac{2005^{2005}+1}{2005^{2006}+1}\)
\(\Rightarrow2005A=\frac{2005^{2006}+2005}{2005^{2006}+1}\)
\(\Rightarrow2005A=1+\frac{2004}{2005^{2006}+1}\)
\(B=\frac{2005^{2004}+1}{2005^{2005}+1}\)
\(\Rightarrow2005B=\frac{2005^{2005}+2005}{2005^{2005}+1}\)
\(\Rightarrow2005B=1+\frac{2004}{2005^{2005}+1}\)
Ta thấy \(\frac{2004}{2005^{2005}+1}>\frac{2004}{2005^{2006}+1}\)
Suy ra \(1+\frac{2004}{2005^{2005}+1}>1+\frac{2004}{2005^{2006}+1}\)
hay 2005B>2005A
Vậy B>A
bn rút gọn đĩog rồi so sánh