Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
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
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 .
\(2005A=\frac{2005\left(2005^{2005}+1\right)}{2005^{2006}+1}=\frac{2005^{2006}+2005}{2005^{2006}+1}=\frac{2005^{2006}+1+2004}{2005^{2006}+1}=\frac{2005^{2006}+1}{2005^{2006}+1}+\frac{2004}{2005^{2006}+1}=1+\frac{2004}{2005^{2006}+1}\)
\(2005B=\frac{2005\left(2005^{2004}+1\right)}{2005^{2005}+1}=\frac{2005^{2005}+2005}{2005^{2005}+1}=\frac{2005^{2005}+1+2014}{2005^{2005}+1}=\frac{2005^{2005}+1}{2005^{2005}+1}+\frac{2014}{2005^{2005}+1}=1+\frac{2014}{2005^{2005}+1}\)Ta thấy \(2005^{2006}+1>2005^{2005}+1\Rightarrow\frac{2004}{2005^{2006}+1}< \frac{2004}{2005^{2005}+1}\Rightarrow1+\frac{2004}{2005^{2006}+1}< 1+\frac{2004}{2005^{2005}+1}\)
\(\Rightarrow A< B\)
a,19/7=5/7 +2
2>7/9 => 19/7>7/9
b, 72/73=1- 1/73
98/99=1- 1/99
1/73>1/99
c,19/18=1+ 1/18
2005/2004=1+ 1/2004
1/18>1/2004
d, 72/73=(58+14)/73=58/73 + 14/73
58/73>58/99
=> 72/73>58/99
Vì \(\frac{2005^{2005}+1}{2005^{2006}+1}\) < 1
Nên \(\frac{2005^{2005}+1}{2005^{2006}+1}\) < \(\frac{2005^{2005}+1+2004}{2005^{2006}+1+2004}\)
Ta có: \(\frac{2005^{2005}+1+2004}{2005^{2006}+1+2004}=\frac{2005^{2005}+2005}{2005^{2006}+2005}=\frac{2005\left(2005^{2004}+1\right)}{2005\left(2005^{2005}+1\right)}=\frac{2005^{2004}+1}{2005^{2005}+1}\)
Nên: \(\frac{2005^{2005}+1}{2005^{2006}+1}\) < \(\frac{2005^{2004}+1}{2005^{2005}+1}\)
=> A < B
\(A=\frac{2005^{2005}+1}{2005^{2006}+1}\) và \(B=\frac{2005^{2004}+1}{2005^{2005}+1}\)
So sánh A và B
\(2005A=\frac{2005^{2005}+1}{2005^{2006}+1}=\frac{2005.\left(2005^{2005}+1\right)}{2005^{2006}+1}=\frac{2005^{2006}+2005}{2005^{2006}}\) \(=\frac{2005^{2006}+2014+1}{2005^{2006}+1}=\frac{2005^{2006}+1}{2005^{2006}+1}+\frac{2004}{2005^{2006}+1}=1+\frac{2004}{2005^{2006}+1}\)
\(2005B=\frac{2005^{2004}+1}{2005^{2005}+1}=\frac{2005.\left(2005^{2004}+1\right)}{2005^{2005}+1}=\frac{2005^{2005}+2005}{2005^{2005}+1}\)\(=\frac{2005^{2005}+2004+1}{2005^{2005}+1}=\frac{2005^{2005}+1}{2005^{2005}+1}+\frac{2004}{2005^{2005}+1}=1+\frac{2004}{2005^{2005}+1}\)
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
sửa chỗ \(\frac{2005^{2006}+2014+1}{2005^{2006}+1}\) thành \(\frac{2005^{2006}+2004+1}{2005^{2006}+1}\)nhé
\(2005A=\frac{2005\left(2005^{2005}+1\right)}{2005^{2006}+1}=\frac{2005^{2006}+2005}{2005^{2006}+1}=\frac{2005^{2006}+1+2004}{2005^{2006}+1}=\frac{2005^{2006}+1}{2005^{2006}+1}+\frac{2004}{2005^{2006}+1}=1+\frac{2004}{2005^{2006}+1}\)
\(2005B=\frac{2005\left(2005^{2004}+1\right)}{2005^{2005}+1}=\frac{2005^{2005}+2005}{2005^{2005}+1}=\frac{2005^{2005}+1+2004}{2005^{2005}+1}=\frac{2005^{2005}+1}{2005^{2005}+1}+\frac{2004}{2005^{2005}+1}=1+\frac{2004}{2005^{2005}+1}\)
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
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
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
\(A=\frac{2005^{2005}+1}{2005^{2006}+1}\)
\(\Rightarrow2005A=\frac{2005^{2006}+2005}{2005^{2006}+1}\)
\(\Rightarrow2005A=1+\frac{2004}{2005^{2006}+1}\)
\(B=\frac{2005^{2004}+1}{2005^{2005}+1}\)
\(\Rightarrow2005B=\frac{2005^{2005}+2005}{2005^{2005}+1}\)
\(\Rightarrow2005B=1+\frac{2004}{2005^{2005}+1}\)
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
Bài 1:19.C=\(\frac{19^{209}+19}{19^{209}+1}\)=\(\frac{19^{209}+1+18}{19^{209}+1}\)=\(\frac{19^{209}+1}{19^{209}+1}\)+\(\frac{18}{19^{209}+1}\)=1+\(\frac{18}{19^{209}+1}\)19D=\(\frac{19^{210}+19}{19^{210}+1}\)=\(\frac{19^{210}+1+18}{19^{210}+1}\)=\(\frac{19^{210}+1}{19^{210}+1}\)+\(\frac{18}{19^{210}+1}\)=1+\(\frac{18}{19^{210}+1}\).Vì \(\frac{18}{19^{209}+1}\)>\(\frac{18}{19^{210}+1}\)nên 19A>19B\(\Rightarrow\)A>B
19D=\(\frac{\left(19^{209}+1\right).19}{19^{210}+1}=\frac{19^{210}+19}{19^{210}+1}=\frac{\left(19^{210}+1\right)+18}{19^{210}+1}=\frac{19^{210}+1}{19^{210}+1}+\frac{18}{19^{210}+1}=1+\frac{18}{19^{210}+1}\)
Vì 19C>19D nên C>D