so sánh A=2005^2005+1/2005^2006+1 với B=2005^2004+1/2005^2005+1
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Những câu hỏi liên quan
PH
0
TN
22 tháng 5 2016
\(2005A=\frac{2005\left(2005^{2005}+1\right)}{2005^{2006}+1}=\frac{2005^{2006}+2005}{2005^{2006}+1}=\frac{2005^{2006}+1+2004}{2005^{2006}+1}=\frac{2005^{2006}+1}{2005^{2006}+1}+\frac{2004}{2005^{2006}+1}=1+\frac{2004}{2005^{2006}+1}\)
\(2005B=\frac{2005\left(2005^{2004}+1\right)}{2005^{2005}+1}=\frac{2005^{2005}+2005}{2005^{2005}+1}=\frac{2005^{2005}+1+2004}{2005^{2005}+1}=\frac{2005^{2005}+1}{2005^{2005}+1}+\frac{2004}{2005^{2005}+1}=1+\frac{2004}{2005^{2005}+1}\)
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
2 tháng 4 2017
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 !!!
PN
3
22 tháng 7 2015
\(A=\frac{2005^{2005}+1}{2005^{2006}+1}\)
\(2005A=\frac{2005^{2006}+2005}{2005^{2006}+1}=\frac{2005^{2006}+1+2004}{2005^{2006}+1}=\frac{2005^{2006}+1}{2005^{2006}+1}+\frac{2004}{2005^{2006}+1}\)
\(B=\frac{2005^{2004}+1}{2005^{2005}+1}\)
\(2005B=\frac{2005^{2005}+2005}{2005^{2005}+1}=\frac{2005^{2005}+1+2004}{2005^{2005}+1}=\frac{2005^{2005}+1}{2005^{2005}+1}+\frac{2004}{2005^{2005}+1}\)
Vì \(\frac{2004}{2005^{2006}+1}
4 tháng 6 2016
a) \(\frac{2004}{2005}=1-\frac{1}{2005}\);\(\frac{2005}{2006}=1-\frac{1}{2006}\)
Vì \(\frac{1}{2005}>\frac{1}{2006}\)=>\(1-\frac{1}{2005}< 1-\frac{1}{2006}\)=>\(\frac{2004}{2005}< \frac{2005}{2006}\)
DL
0
HL
1
25 tháng 6 2015
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
bn rút gọn đĩog rồi so sánh