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20 tháng 7 2017

Ta có : \(\frac{2008}{\sqrt{2009}}+\frac{2009}{\sqrt{2008}}=\frac{2009-1}{\sqrt{2009}}+\frac{2008+1}{\sqrt{2008}}=\sqrt{2009}+\sqrt{2008}+\left(\frac{1}{\sqrt{2008}}-\frac{1}{\sqrt{2009}}\right)\)

Vì \(\frac{1}{\sqrt{2008}}>\frac{1}{\sqrt{2009}}\) nên \(\frac{1}{\sqrt{2008}}-\frac{1}{\sqrt{2009}}>0\)

\(\Rightarrow\sqrt{2009}+\sqrt{2008}+\left(\frac{1}{\sqrt{2008}}-\frac{1}{\sqrt{2009}}\right)>\sqrt{2009}+\sqrt{2008}\)

Hay \(\frac{2008}{\sqrt{2009}}+\frac{2009}{\sqrt{2008}}>\sqrt{2008}+\sqrt{2009}\)

21 tháng 7 2017

Cảm ơn bạn CTV 

26 tháng 2 2020

Ta có :

\(\frac{2008}{\sqrt{2009}}+\frac{2009}{\sqrt{2008}}=\frac{2009}{\sqrt{2009}}-\frac{1}{\sqrt{2009}}+\frac{2008}{\sqrt{2008}}+\frac{1}{\sqrt{2008}}\)

\(=\sqrt{2008}+\sqrt{2009}+\frac{1}{\sqrt{2008}}-\frac{1}{\sqrt{2009}}\)

\(\sqrt{2008}< \sqrt{2009}\Rightarrow\frac{1}{\sqrt{2008}}>\frac{1}{\sqrt{2009}}\Leftrightarrow\frac{1}{\sqrt{2008}}-\frac{1}{\sqrt{2009}}\)

\(\Leftrightarrow\sqrt{2008}+\sqrt{2009}+\frac{1}{\sqrt{2008}}-\frac{1}{\sqrt{2009}}>\sqrt{2008}+\sqrt{2009}\)

⇒ đpcm

5 tháng 2 2020

Ta có : \(\frac{2008}{\sqrt{2009}}+\frac{2009}{\sqrt{2008}}\) = \(\frac{2009-1}{\sqrt{2009}}+\frac{2008+1}{\sqrt{2008}}\)

= \(\frac{2009}{\sqrt{2009}}-\frac{1}{\sqrt{2009}}+\frac{2008}{\sqrt{2008}}+\frac{1}{\sqrt{2008}}\)

= \(\frac{\left(\sqrt{2009}\right)^2}{\sqrt{2009}}-\frac{1}{\sqrt{2009}}+\frac{\left(\sqrt{2008}\right)^2}{\sqrt{2008}}+\frac{1}{\sqrt{2008}}\)

= \(\sqrt{2009}-\frac{1}{\sqrt{2009}}+\sqrt{2008}+\frac{1}{\sqrt{2008}}\)

\(\frac{1}{\sqrt{2008}}>\frac{1}{\sqrt{2009}}\)

=> \(\frac{1}{\sqrt{2008}}-\frac{1}{\sqrt{2009}}>0\)

=> \(\frac{1}{\sqrt{2008}}-\frac{1}{\sqrt{2009}}+\sqrt{2008}+\sqrt{2009}>\sqrt{2008}+\sqrt{2009}\)

Vậy \(\frac{1}{\sqrt{2008}}-\frac{1}{\sqrt{2009}}+\sqrt{2008}+\sqrt{2009}>\sqrt{2008}+\sqrt{2009}\) .

\(C=\frac{\frac{1}{2008}-\frac{1}{2009}-\frac{1}{2010}}{\frac{5}{2008}-\frac{5}{2009}-\frac{5}{2010}}+\frac{\frac{2}{2007}-\frac{2}{2008}-\frac{2}{2009}}{\frac{3}{2007}-\frac{3}{2008}-\frac{3}{2009}}\)

\(=\frac{\frac{1}{2008}-\frac{1}{2009}-\frac{1}{2010}}{5.\left(\frac{1}{2008}-\frac{1}{2009}-\frac{1}{2010}\right)}+\frac{2.\left(\frac{1}{2007}-\frac{1}{2008}-\frac{1}{2009}\right)}{3.\left(\frac{1}{2007}-\frac{1}{2008}-\frac{1}{2009}\right)}\)

\(=\frac{1}{5}+\frac{2}{3}\)

\(=\frac{13}{15}\)

24 tháng 8 2016

Ta có : \(\frac{a_1}{a_2}=\frac{a_2}{a_3}=\frac{a_3}{a_4}=...=\frac{a_{2008}}{a_{2009}}=\frac{a_1+a_2+a_3+...+a_{2008}}{a_2+a_3+a_4+...+a_{2009}}\)

Đặt \(\frac{a_1+a_2+a_3+...+a_{2008}}{a_2+a_3+a_4+...+a_{2009}}=b\)thì \(\frac{a_1}{a_2}=b\left(1\right);\frac{a_2}{a_3}=b\left(2\right);\frac{a_3}{a_4}=b\left(3\right);...;\frac{a_{2008}}{a_{2009}}=b\left(2008\right)\)

Nhân (1),(2),(3),...,(2008) vế theo vế,ta có :

 \(\frac{a_1}{a_2}.\frac{a_2}{a_3}.\frac{a_3}{a_4}.....\frac{a_{2008}}{a_{2009}}=b^{2008}\)hay \(\frac{a_1}{a_{2009}}=\left(\frac{a_1+a_2+a_3+...+a_{2008}}{a_2+a_3+a_4+...+a_{2009}}\right)^{2008}\)(đpcm)

10 tháng 10 2015

\(\frac{a_1}{a_2}=\frac{a_2}{a_3}=....=\frac{a_{2008}}{a_{2009}}=\frac{a_1+a_2+....+a_{2008}}{a_2+a_3+....+a_{2009}}\)

=> \(\left(\frac{a_1}{a_2}\right)^{2008}=\left(\frac{a_2}{a_3}\right)^{2008}=....=\left(\frac{a_{2008}}{a_{2009}}\right)^{2008}=\left(\frac{a_1+a_2+....+a_{2008}}{a_2+a_3+....+a_{2009}}\right)^{2008}\)

\(=\frac{a_1.a_2....a_{2008}}{a_2.a_3....a_{2009}}=\frac{a_1}{a_{2009}}\)

=> \(\left(\frac{a_1+a_2+....+a_{2008}}{a_2+a_3+....+a_{2009}}\right)^{2008}=\frac{a_1}{a_{2009}}\)

=> Đpcm

 

26 tháng 7 2017

ta có: \(A=\frac{2006}{2007}+\frac{2007}{2008}+\frac{2008}{2009}+\frac{2009}{2006}\)

A = \(1-\frac{1}{2007}+1-\frac{1}{2008}+1-\frac{1}{2009}+1+\frac{3}{2006}\)

A= \(4\)\(+\frac{3}{2006}-\left(\frac{1}{2007}+\frac{1}{2008}+\frac{1}{2009}\right)\)

Do 1/2007 < 1/2006 ; 1/2008<1/2006 ; 1/2009<1/2006=> 1/2007 + 1/2008 + 1/2009 < 1/2006 + 1/2006 + 1/2006

Mà 1/2006 + 1/2006 + 1/2006 = 3/2006

=> 3/2006  -( 1/2007 + 1/2008 + 1/2009) > 0 

=> \(4+\frac{3}{2006}-\left(\frac{1}{2007}+\frac{1}{2008}+\frac{1}{2009}\right)>4\)

=> A > 4

26 tháng 7 2017

Ta có:\(\frac{2006}{2007}< 1\)

           \(\frac{2007}{2008}< 1\)

           \(\frac{2008}{2009}< 1\)

            \(\frac{2009}{2006}>1\)\(\frac{2006}{2007}+\frac{2007}{2008}+\frac{2008}{2009}+\frac{2009}{2006}< 4\)