DH
2
K
Khách
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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 \(\frac{2004}{2005^{2006}+1}<\frac{2004}{2005^{2005}+1}\)
Nên A<B
PH
0
DL
0
NV
3
TN
22 tháng 3 2016
\(\Leftrightarrow10A=\frac{10\left(10^{2004}+1\right)}{10^{2005}+1}\)
\(\Rightarrow10A=\frac{10^{2005}+10}{10^{2005}+1}\)
\(10A=\frac{10^{2005}+1+9}{10^{2005}+1}=\frac{10^{2005}+1}{10^{2005}+1}+\frac{9}{10^{2005}+1}\)
\(10A=1+\frac{9}{10^{2005}+1}\)
tương tự như trên ta có :
\(10B=1+\frac{9}{10^{2006}+1}\)
ta thấy:102005+1<102006+1
\(\Rightarrow\frac{9}{10^{2005}+1}>\frac{9}{10^{2006}+1}\)
\(\Rightarrow1+\frac{9}{10^{2005}+1}>1+\frac{9}{10^{2006}+1}\)
=>10A>10B
=>A>B
kl: vậy A>B
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
LT
0
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 !!!
24 tháng 7 2018
mik ko bít
I don't now
................................
.............
x+2/3=-1/12*-4/5; 2) x-1/4=5/8*2/3; 3) 2/15*x=2/7; 4) x:3/-4=-|12|; 5) x/4=-3/6; 6) x/7=6/21; 7) x/3=19/8+55/24; 8) 2/x=5/7+-1/21
\(A=\frac{4^{2005}+1}{4^{2006}+1},B=\frac{4^{2004}+1}{4^{2005}+1}\)
\(\frac{1}{A}=\frac{4^{2006}+1}{4^{2005}+1}=\frac{4^{2006}+4}{4^{2005}+1}-\frac{3}{4^{2005}+1}=4-\frac{3}{4^{2005}+1}\)
\(\frac{1}{B}=\frac{4^{2005}+1}{4^{2004}+1}=\frac{4^{2005}+4}{4^{2004}+4}-\frac{3}{4^{2004}+1}=4-\frac{3}{4^{2004}+1}\)
Có \(4^{2005}+1>4^{2004}+1>0\Rightarrow\frac{3}{4^{2005}+1}< \frac{3}{4^{2004}+1}\Rightarrow\frac{-3}{4^{2005}+1}>-\frac{3}{4^{2004}+1}\)
Suy ra \(\frac{1}{A}>\frac{1}{B}\Rightarrow A< B\).