a)so sánh :920 và 2713 . Vì sao?
b) so sánh : A=\(\frac{2005^{2005}+1}{20005^{2006}+1}vàB=\frac{2005^{2004}+1}{2005^{2005}+1}\)vì sao?
K
Khách
Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
Những câu hỏi liên quan
4 tháng 6 2016
a) \(\frac{2004}{2005}=1-\frac{1}{2005}\);\(\frac{2005}{2006}=1-\frac{1}{2006}\)
Vì \(\frac{1}{2005}>\frac{1}{2006}\)=>\(1-\frac{1}{2005}< 1-\frac{1}{2006}\)=>\(\frac{2004}{2005}< \frac{2005}{2006}\)
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 \(2005^{2006}+1>2005^{2005}+1\Rightarrow\frac{2004}{2005^{2006}+1}< \frac{2004}{2005^{2005}+1}\Rightarrow1+\frac{2004}{2005^{2006}+1}< 1+\frac{2004}{2005^{2005}+1}\)
\(\Rightarrow A< B\)
KY
\(A=\frac{2005^{2005}+1}{2005^{2006}+1}\) và \(B=\frac{2005^{2004}+1}{2005^{2005}+1}\)
So sánh A và B
2
29 tháng 5 2018
\(2005A=\frac{2005^{2005}+1}{2005^{2006}+1}=\frac{2005.\left(2005^{2005}+1\right)}{2005^{2006}+1}=\frac{2005^{2006}+2005}{2005^{2006}}\) \(=\frac{2005^{2006}+2014+1}{2005^{2006}+1}=\frac{2005^{2006}+1}{2005^{2006}+1}+\frac{2004}{2005^{2006}+1}=1+\frac{2004}{2005^{2006}+1}\)
\(2005B=\frac{2005^{2004}+1}{2005^{2005}+1}=\frac{2005.\left(2005^{2004}+1\right)}{2005^{2005}+1}=\frac{2005^{2005}+2005}{2005^{2005}+1}\)\(=\frac{2005^{2005}+2004+1}{2005^{2005}+1}=\frac{2005^{2005}+1}{2005^{2005}+1}+\frac{2004}{2005^{2005}+1}=1+\frac{2004}{2005^{2005}+1}\)
Vì \(2005^{2006}+1>2005^{2005}+1\)
Nên \(1+\frac{2004}{2005^{2006}+1}< 1+\frac{2004}{2005^{2005}+1}\)
Hay A < B
Vậy A < B
29 tháng 5 2018
sửa chỗ \(\frac{2005^{2006}+2014+1}{2005^{2006}+1}\) thành \(\frac{2005^{2006}+2004+1}{2005^{2006}+1}\)nhé
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
NV
2
9 tháng 1 2016
Nhân a và b với 2005 ta có : 2005.a =\(\frac{2005.\left(2005^{2005}+1\right)}{2005^{2006}+1}\)=\(\frac{2005^{2006}+2005}{2005^{2006}+1}\)= \(\frac{\left(2005^{2006}+1\right)+2004}{2005^{2006}+1}\)= \(\frac{2005^{2006}+1}{2005^{2006}+1}\)+ \(\frac{2004}{2005^{2006}+1}\)=1+\(\frac{2004}{2005^{2006}+1}\) 2005.b = \(\frac{2005.\left(2005^{2004}+1\right)}{2005^{2005}+1}\)=\(\frac{2005^{2005}+2005}{2005^{2005}+1}\)= \(\frac{\left(2005^{2005}+1\right)+2004}{2005^{2005}+1}\)=\(\frac{2005^{2005}+1}{2005^{2005}+1}\)+ \(\frac{2004}{2005^{2005}+1}\) =1+\(\frac{2004}{2005^{2005}+1}\) Vì 2004=2004 , 2005^2005 +1 < 2005^2006 + 1 => \(\frac{2004}{2005^{2006}+1}\)< \(\frac{2004}{2005^{2005}+1}\)=> a<b Vậy A < B
VD
9 tháng 1 2016
B=(2005(2005^2004+1))/(2005(2005^2005+1))=(2005^2005+2005)/(2005^2006+2005)
Có 1-A=(2005^2006-2005^2005)/(2005^2006+1)
1-B=(2005^2006-2005^2005)/(2005^2006+2005)
suy ra 1-A>1-B.Suy ra 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
3 tháng 5 2017
\(10A=\frac{2005^{2006}+10}{2005^{2006}+1}\)
\(10B=\frac{2005^{2005}+10}{2005^{2005}+1}\)
Rồi bạn so sánh 10A và 10B là ra.
Ai thấy đúng thì ủng hộ nha !!!, sai thì góp ý cho mink nha
LV
3 tháng 5 2017
Ta có
A <\(\frac{2005^{2005}+2005}{2005^{2006}+2005}=\frac{2005\left(2005^{2004}+1\right)}{2005\left(2005^{2005}+1\right)}\)=\(\frac{2005^{2004}+1}{2005^{2005}+1}\)
\(\RightarrowĐPCM\)
a ) Ta có : \(9^{20}\)= \(\left(3^2\right)^{10}\)= \(3^{20}\)
\(27^{13}\)= \(\left(3^3\right)^{13}\)= \(3^{39}\)
Vì 39 > 20 => 9^ 20 < 27 ^ 13
Phần b bạn vào câu hỏi tương tự. Nhớ tích đúng cho tớ