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Ta có B=\(\frac{2009^{2010}-2}{2009^{2011}-2}\)<1
=>\(\frac{2009^{2010}-2}{2009^{2011}-2}\)<\(\frac{2009^{2010}-2+3}{2009^{2011}-2+3}\)=\(\frac{2009^{2010}+1}{2009^{2011}+1}\)(1)
Mà \(\frac{2009^{2010}+1}{2009^{2011}+1}\)<1
=> \(\frac{2009^{2010}+1}{2009^{2011}+1}\)<\(\frac{2009^{2010}+1+2008}{2009^{2011}+1+2008}\)=\(\frac{2009^{2010}+2009}{2009^{2011}+2009}\)=\(\frac{2009\cdot\left(2009^{2009}+1\right)}{2009\cdot\left(2009^{2010}+1\right)}\)=\(\frac{2009^{2009}+1}{2009^{2010}+1}\)=A(2)
Từ (1)và(2)=>B<\(\frac{2009^{2010}+1}{2009^{2011}+1}\)<A=>B<A hay A>B
\(\dfrac{123}{456}\cdot\left(\dfrac{2010}{2011}-\dfrac{2011}{2010}\right)-\left(\dfrac{2009}{2010}-\dfrac{1}{2011}\right):\dfrac{456}{123}\)
\(=\dfrac{123}{456}\cdot\left(\dfrac{2010}{2011}-\dfrac{2011}{2010}\right)-\left(\dfrac{2009}{2010}-\dfrac{1}{2011}\right)\cdot\dfrac{123}{456}\)
\(=\dfrac{123}{456}\left[\left(\dfrac{2010}{2011}-\dfrac{2011}{2010}\right)-\left(\dfrac{2009}{2010}-\dfrac{1}{2011}\right)\right]\)
\(=\dfrac{123}{456}\left(\dfrac{2010}{2011}-\dfrac{2011}{2010}-\dfrac{2009}{2010}+\dfrac{1}{2011}\right)\)
\(=\dfrac{123}{456}\left[\left(\dfrac{2010}{2011}+\dfrac{1}{2011}\right)-\left(\dfrac{2011}{2010}+\dfrac{2009}{2010}\right)\right]\)
\(=\dfrac{123}{456}\left(1-2\right)\)
\(=-\dfrac{123}{456}\)
C=(1-2-3+4)+(5-6-7+8)+...+(2005-2006-2007+2008)+2009-2010-2011
=-1-2011
=-2012
Đặt \(\left\{{}\begin{matrix}x-2008=n\\2x+2009=h\\3x-2011=t\end{matrix}\right.\Rightarrow n+h+t=6x-2010\)
\(\Rightarrow pt\Leftrightarrow\dfrac{1}{n}+\dfrac{1}{h}=\dfrac{1}{n+h+t}-\dfrac{1}{t}\)
\(\Leftrightarrow\dfrac{n+h}{hn}=\dfrac{-\left(n+h\right)}{t\left(n+h+t\right)}\)
\(\Leftrightarrow\left(n+h\right)\left(\dfrac{1}{hn}+\dfrac{1}{t\left(n+h+t\right)}\right)=0\)
\(\Leftrightarrow\left(n+h\right)\dfrac{t\left(n+h+t\right)+hn}{hnt\left(n+h+t\right)}=0\)
\(\Leftrightarrow\dfrac{\left(n+h\right)\left(n+t\right)\left(t+h\right)}{hnt\left(n+h+t\right)}=0\)
\(\Rightarrow\left[{}\begin{matrix}n=-h\\n=-t\\t=-h\end{matrix}\right.\)\(\Rightarrow\left[{}\begin{matrix}x-2008=-\left(2x+2009\right)\\x-2008=-\left(3x-2011\right)\\3x-2011=-\left(2x+2009\right)\end{matrix}\right.\)\(\Rightarrow\left[{}\begin{matrix}x=-\dfrac{1}{3}\\x=\dfrac{4019}{4}\\x=\dfrac{2}{5}\end{matrix}\right.\)
Có \(\frac{20092009}{20102010}=\frac{2009.10001}{2010.10001}=\frac{2009}{2010}\)
=> \(\frac{2009}{2010}=\frac{20092009}{20102010}\)
\(A=\dfrac{2010}{2}+\dfrac{2010}{6}+...+\dfrac{2010}{9900}\\ =2010\left(\dfrac{1}{2}+\dfrac{1}{6}+\dfrac{1}{12}+...+\dfrac{1}{9900}\right)\\ =2010\left(\dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+...+\dfrac{1}{99.100}\right)\\ =2010\left(\dfrac{1}{1}-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{99}-\dfrac{1}{100}\right)\\ =2010.\dfrac{99}{100}=\dfrac{19899}{10}\)
Ta có:
\(A=1+2+2^2+2^3+...+2^{2010}\)
\(2A=2+2^2+2^3+2^4+...+2^{2011}\)
\(\Rightarrow2A-A=\left(2+2^2+2^3+...+2^{2011}\right)-\left(1+2+2^2+...+2^{2010}\right)\)
\(\Rightarrow A=2^{2011}-1\)
\(B=2^{2011}-1\)
Vậy A = B.
