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\(A=\frac{2005^{2005}+1}{2005^{2006}+1}\)
\(\Rightarrow2005A=\frac{2005^{2006}+2005}{2005^{2006}+1}\)
\(\Rightarrow2005A=1+\frac{2004}{2005^{2006}+1}\)
\(B=\frac{2005^{2004}+1}{2005^{2005}+1}\)
\(\Rightarrow2005B=\frac{2005^{2005}+2005}{2005^{2005}+1}\)
\(\Rightarrow2005B=1+\frac{2004}{2005^{2005}+1}\)
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
a, Ta có: \(\frac{2012.2013}{2012.2013+1}< 1< \frac{2013}{2012}\)
\(\Rightarrow\frac{2012.2013}{2012.2013+1}< \frac{2013}{2012}\)
b, \(A=\frac{2003.2004-1}{2003.2004}=1-\frac{1}{2003.2004}\)
\(B=\frac{2004.2005-1}{2004.2005}=1-\frac{1}{2004.2005}\)
Ta có: \(2003.2004< 2004.2005\)
\(\Rightarrow\frac{1}{2003.2004}>\frac{1}{2004.2005}\)
\(\Rightarrow1-\frac{1}{2003.2004}< 1-\frac{1}{2004.2005}\)
\(\Rightarrow A< B\)
a) Vì A=\(\dfrac{15^{16}+1}{15^{17}+1}\) < 1
\(\Rightarrow\dfrac{15^{16}+1}{15^{17}+1}< \dfrac{15^{16}+1+14}{15^{17}+1+14}=\dfrac{15^{16}+15}{15^{17}+15}\) \(=\dfrac{15\left(15^{15}+1\right)}{15\left(15^{16}+1\right)}\) \(=\dfrac{15^{15}+1}{15^{16}+1}\)
Vậy A<B
b) A=\(\dfrac{2006^{2007}+1}{2006^{2006}+1}>1\)
\(\Rightarrow\dfrac{2006^{2007}+1+2005}{2006^{2006}+1+2005}\)
= \(\dfrac{2006^{2007}+2006}{2006^{2006}+2006}\)
= \(\dfrac{2006\left(2006^{2006}+1\right)}{2006\left(2006^{2005}+1\right)}\)
= \(\dfrac{2006^{2006+1}}{2006^{2005}+1}\)
Vậy A>B
A= (1000+1)x1000:2=500500
B=1.2.3.4........11=39916800
=>A<B
tich ủng hộ nha
2006.A = 2006. 20062006 + 1 = 20062007 + 2006 = 20062007 + 1 + 2005 = 20062007 + 1 + 2005 = 1 + 2005
20062007 + 1 20062007 + 1 20062007 + 1 20062007 + 1 20062007 + 1 20062007 + 1 Sau đó,với phần B bạn làm tương tự thì sẽ ra A > B
Bạn có thể tham khảo ở đây :
Câu hỏi của Vân Trang Bùi - Toán lớp 6 | Học trực tuyến
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 !!!
\(\Leftrightarrow10A=\frac{10\left(10^{2004}+1\right)}{10^{2005}+1}\)
\(\Rightarrow10A=\frac{10^{2005}+10}{10^{2005}+1}\)
\(10A=\frac{10^{2005}+1+9}{10^{2005}+1}=\frac{10^{2005}+1}{10^{2005}+1}+\frac{9}{10^{2005}+1}\)
\(10A=1+\frac{9}{10^{2005}+1}\)
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
\(A=\frac{1000^{2004}+1}{1000^{2005}+1}\)
=> \(1000A=\frac{1000^{2005}+1000}{1000^{2005}+1}=1+\frac{999}{1000^{2005}+1}\)
\(B=\frac{1000^{2005}+1}{1000^{2006}+1}\)
=> \(1000A=\frac{1000^{2006}+1000}{1000^{2006}+1}=1+\frac{999}{1000^{2006}+1}\)
Vì: \(1000^{2006}+1>1000^{2005}+1\)
=> \(\frac{999}{1000^{2006}+1}< \frac{99}{1000^{2005}+1}\)
=> \(1+\frac{999}{1000^{2006}+1}< 1+\frac{99}{1000^{2005}+1}\)
=> 1000B < 1000A
=> B < A