Bài 1: Chứng tỏ rằng:(146 - 493) \(⋮\) 63
Bài 2: So sánh: \(\frac{2004}{2005}\)+ \(\frac{2005}{2006}\)+\(\frac{2005}{2004}\) với 3
K
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Những câu hỏi liên quan
PD
2
HT
8 tháng 5 2016
Ta thấy :
2004/2005 >2004/2005+2006
2005/2006> 2005/2005+2006
=> 2004/2005 + 2005/2006 >2004+2005 / 2005+2006
8 tháng 5 2016
Ta có: \(\frac{2004}{2005}>\frac{2004}{2005+2006}\) (1)
\(\frac{2005}{2006}>\frac{2005}{2005+2006}\) (2)
Từ (1) và (2) => \(\frac{2004}{2005}+\frac{2005}{2006}>\frac{2004+2005}{2005+2006}\) => M>N
Ai k mik mik k lại. chúc các bạn thi tốt
15 tháng 3 2018
Ta có :
\(B=\frac{2004+2005}{2005+2006}=\frac{2004}{2005+2006}+\frac{2005}{2005+2006}< \frac{2004}{2005}+\frac{2005}{2006}=A\)
\(\Rightarrow\)\(B< A\) hay \(A>B\)
Vây \(A>B\)
Chúc bạn học tốt ~
13 tháng 6 2018
\(A=\frac{2005^{2005}+1}{2005^{2006}+1}\)
\(\Rightarrow2005A=\frac{2005^{2006}+2005}{2005^{2006}+1}\)
\(\Rightarrow2005A=1+\frac{2004}{2005^{2006}+1}\)
\(B=\frac{2005^{2004}+1}{2005^{2005}+1}\)
\(\Rightarrow2005B=\frac{2005^{2005}+2005}{2005^{2005}+1}\)
\(\Rightarrow2005B=1+\frac{2004}{2005^{2005}+1}\)
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
23 tháng 5 2018
Ta có VẾ A
\(A=\frac{2005^{2005}+1}{2005^{2006}+1}\)
\(2005\cdot A=\frac{2005\cdot\left(2005^{2005}+1\right)}{2005^{2006}+1}\)
\(2005\cdot A=\frac{2005^{2006}+2005}{2005^{2006}+1}\)
\(2005\cdot A=\frac{2005^{2006}+1+2004}{2005^{2006}+1}\)
\(2005\cdot A=1+\frac{2004}{2005^{2006}+1}\)
Ta lại có Vế B :
\(B=\frac{2005^{2004}+1}{2005^{2005}+1}\)
\(2005\cdot B=\frac{2005\cdot\left(2005^{2004}+1\right)}{2005^{2005}+1}\)
\(2005\cdot B=\frac{2005^{2005}+2005}{2005^{2005}+1}\)
\(2005\cdot B=\frac{2005^{2005}+1+2004}{2005^{2005}+1}\)
\(2005\cdot B=1+\frac{2004}{2005^{2005}+1}\)
Nhìn vào trên , suy ra A < B .
WH
23 tháng 5 2018
\(2005A=\frac{2005\left(2005^{2005}+1\right)}{2005^{2006}+1}=\frac{2005^{2006}+2005}{2005^{2006}+1}=\frac{2005^{2006}+1+2004}{2005^{2006}+1}=\frac{2005^{2006}+1}{2005^{2006}+1}+\frac{2004}{2005^{2006}+1}=1+\frac{2004}{2005^{2006}+1}\)
\(2005B=\frac{2005\left(2005^{2004}+1\right)}{2005^{2005}+1}=\frac{2005^{2005}+2005}{2005^{2005}+1}=\frac{2005^{2005}+1+2014}{2005^{2005}+1}=\frac{2005^{2005}+1}{2005^{2005}+1}+\frac{2014}{2005^{2005}+1}=1+\frac{2014}{2005^{2005}+1}\)Ta thấy \(2005^{2006}+1>2005^{2005}+1\Rightarrow\frac{2004}{2005^{2006}+1}< \frac{2004}{2005^{2005}+1}\Rightarrow1+\frac{2004}{2005^{2006}+1}< 1+\frac{2004}{2005^{2005}+1}\)
\(\Rightarrow A< B\)
HT
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
SY
1 tháng 4 2016
A=\(\frac{2005^{2005}+1}{2005^{2006}+1}\) < 1 => \(\frac{2005^{2005}+1}{2005^{2006}+1}\) < \(\frac{2005^{2005}+1+2004}{2005^{2006}+1+2004}\) = \(\frac{2005^{2005}+2005}{2005^{2006}+2005}\)= \(\frac{2005.\left(2005^{2004}+1\right)}{2005.\left(2005^{2005}+1\right)}\) = \(\frac{2005^{2004}+1}{2005^{2005}+1}\) = B => A<B.
TV
3
XO
21 tháng 1 2020
Ta có : \(2005C=\frac{2005\left(2005^{2005}+1\right)}{2005^{2006}+1}=\frac{2005^{2006}+1+2004}{2005^{2006}+1}=1+\frac{2004}{2005^{2006}+1}\)
\(2005D=\frac{2005\left(2005^{2004}+1\right)}{2005^{2005}+1}=\frac{2005^{2005}+1+2004}{2005^{2005}+1}=1+\frac{2004}{2005^{2005}+1}\)
Vì \(\frac{2004}{2005^{2006}+1}< \frac{2004}{2005^{2005}+1}\Rightarrow1+\frac{2004}{2005^{2006}+1}< 1+\frac{2004}{2005^{2005}+1}\)
=> 2005.C < 2005.D
=> C < D