So Sánh
a,A= \(\frac{2008^{2008}+1}{2008^{2009}+1}\)và B=\(\frac{2008^{2007}+1}{2008^{2008}+1}\)
b, M=\(\frac{100^{100}+1}{100^{99}+1}\)và N= \(\frac{100^{101}+1}{100^{100}+1}\)
K
Khách
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PA
1
22 tháng 1 2019
Xin chào các bạn !!!
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26 tháng 5 2018
1.
\(1+\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^{99}}+\frac{1}{2^{100}}+\frac{1}{2^{100}}\)
\(=1+\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^{99}}+\left(\frac{1}{2^{100}}+\frac{1}{2^{100}}\right)\)
\(=1+\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^{99}}+\frac{1}{2^{99}}\)
cứ làm như vậy ta được :
\(=1+1=2\)
26 tháng 5 2018
2. Ta có :
\(\frac{2008+2009}{2009+2010}=\frac{2008}{2009+2010}+\frac{2009}{2009+2010}\)
vì \(\frac{2008}{2009}>\frac{2008}{2009+2010}\); \(\frac{2009}{2010}>\frac{2009}{2009+2010}\)
\(\Rightarrow\frac{2008}{2009}+\frac{2009}{2010}>\frac{2008+2009}{2009+2010}\)
4 tháng 5 2017
\(A=\frac{100^{2007}+1}{100^{2008}+1}\Rightarrow100.A=\frac{100^{2008}+100}{100^{2008}+1}=\frac{100^{2008}+1+99}{100^{2008}+1}=1+\frac{99}{100^{2008}+1}\)
\(B=\frac{100^{2006}+1}{100^{2007}+1}\Rightarrow100.B=\frac{100^{2007}+100}{100^{2007}+1}=\frac{100^{2007}+1+99}{100^{2007}+1}=1+\frac{99}{100^{2007}+1}\)
Vì \(\frac{99}{100^{2007}+1}>\frac{99}{100^{2008}+1};1=1\Rightarrow1+\frac{99}{100^{2007}+1}>1+\frac{99}{100^{2008}+1}\)hay \(A>B\)
Vậy \(A>B\)
DO
2
NH
20 tháng 12 2019
ý, nếu không được dùng cách kia thì làm cách này cho chắc đi :v
Ta có: \(2008A=\frac{2008\left(2008^{2008}+1\right)}{2008^{2009}+1}=\frac{2008^{2009}+2008}{2008^{2009}+1}=\frac{\left(2008^{2009}+1\right)+2007}{2008^{2009}+1}=1+\frac{2007}{2008^{2009}+1}\)
Lại có: \(2008B=\frac{2008\left(2008^{2007}+1\right)}{2008^{2008}+1}=\frac{2008^{2008}+2008}{2008^{2008}+1}=\frac{\left(2008^{2008}+1\right)+2007}{2008^{2008}+1}=1+\frac{2007}{2008^{2008}+1}\)
Vì 2008 < 2009 \(\Rightarrow2008^{2008}< 2008^{2009}\)\(\Rightarrow2008^{2008}+1< 2008^{2009}+1\)\(\Rightarrow\frac{2007}{2008^{2008}+1}>\frac{2007}{2008^{2009}+1}\)\(\Rightarrow1+\frac{2007}{2008^{2008}+1}>1+\frac{2007}{2008^{2009}+1}\)\(\Rightarrow2008B>2008A\)\(\Rightarrow B>A\)
NH
20 tháng 12 2019
Vì A <1 , B < 1
Nên ta có: \(A=\frac{2008^{2008}+1}{2008^{2009}+1}< \frac{2008^{2008}+1+2007}{2008^{2009}+1+2007}=\frac{2008^{2008}+2008}{2008^{2009}+2008}=\frac{2008\left(2008^{2007}+1\right)}{2008\left(2008^{2008}+1\right)}=\frac{2008^{2007}+1}{2008^{2008}+1}=B\)
7 tháng 9 2017
A=\(\frac{2007^{2007}}{2008^{2008}}\)
B=\(\frac{2008^{2008}}{2009^{2009}}\)
a) Áp dụng \(\frac{a}{b}< 1\Leftrightarrow\frac{a}{b}< \frac{a+m}{b+m}\) (a;b;m \(\in\) N*)
Ta có:
\(A=\frac{2008^{2008}+1}{2008^{2009}+1}< \frac{2008^{2008}+1+2007}{2009^{2009}+1+2007}\)
\(A< \frac{2008^{2008}+2008}{2008^{2009}+2008}\)
\(A< \frac{2008.\left(2008^{2007}+1\right)}{2008.\left(2008^{2008}+1\right)}=\frac{2008^{2007}+1}{2008^{2008}+1}=B\)
=> A < B
b) Áp dụng \(\frac{a}{b}>1\Leftrightarrow\frac{a}{b}>\frac{a+m}{b+m}\) (a;b;m \(\in\) N*)
Ta có:
\(N=\frac{100^{101}+1}{100^{100}+1}>\frac{100^{101}+1+99}{100^{100}+1+99}\)
\(N>\frac{100^{101}+100}{100^{100}+100}\)
\(N>\frac{100.\left(100^{100}+1\right)}{100.\left(100^{99}+1\right)}=\frac{100^{100}+1}{100^{99}+1}=M\)
=> M > N
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