BT7: So sánh
1) \(A=\dfrac{2005^{2005}+1}{2005^{2006}+1}\)và \(B=\dfrac{2005^{2004}+1}{2005^{2005}+1}\)
K
Khách
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Những câu hỏi liên quan
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}
PH
0
4 tháng 6 2016
a) \(\frac{2004}{2005}=1-\frac{1}{2005}\);\(\frac{2005}{2006}=1-\frac{1}{2006}\)
Vì \(\frac{1}{2005}>\frac{1}{2006}\)=>\(1-\frac{1}{2005}< 1-\frac{1}{2006}\)=>\(\frac{2004}{2005}< \frac{2005}{2006}\)
DL
0
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
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 \(2005^{2006}+1>2005^{2005}+1\Rightarrow\frac{2004}{2005^{2006}+1}< \frac{2004}{2005^{2005}+1}\Rightarrow1+\frac{2004}{2005^{2006}+1}< 1+\frac{2004}{2005^{2005}+1}\)
\(\Rightarrow A< B\)
LT
0
Nếu:
\(\dfrac{a}{b}< 1\Rightarrow\dfrac{a+m}{b+m}< 1\left(m\in N\right)\)
\(A=\dfrac{2005^{2005}+1}{2005^{2006}+1}< 1\)
\(A< \dfrac{2005^{2005}+1+2004}{2005^{2006}+1+2004}\Rightarrow A< \dfrac{2005^{2005}+2005}{2005^{2006}+2005}\Rightarrow A< \dfrac{2005\left(2005^{2004}+1\right)}{2005\left(2005^{2005}+1\right)}\Rightarrow A< \dfrac{2005^{2004}+1}{2005^{2005}+1}=B\)
\(A< B\)
Ta có : A = \(\dfrac{2005^{2005}+1}{2005^{2006}+1}\)
\(2005\)A = \(\dfrac{\left(2005^{2005}+1\right).2005}{2005^{2006}+1}\)
\(2005\)\(A\)= \(\dfrac{2005^{2006}+2005}{2005^{2006}+1}\)
\(2005\)\(A\)= \(\dfrac{2005^{2006}+1+2004}{2005^{2006}+1}\)
\(2005A=\dfrac{2005^{2006}+1}{2005^{2006}+1}+\dfrac{2004}{2005^{2006}+1}\)
\(2005A=1+\dfrac{2004}{2005^{2006}+1}\)
Tương tự như vậy với \(B\) ta đc
\(2005B=1+\dfrac{2004}{2005^{2005}+1}\)
Vì \(2005^{2006}+1>2005^{2005}+1\)
\(=>\) \(1+\dfrac{2004}{2005^{2006}+1}\)\(< \)\(1+\dfrac{2004}{2005^{2005}+1}\)
\(=>\)\(2005A< 2005B\)
\(=>\)\(A< B\)
Vậy \(A< B\)