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\(\frac{x-2009-2010}{2008}+\frac{x-2008-2010}{2009}+\frac{x-2008-2009}{2010}=3\)
\(\Leftrightarrow\frac{x-2008-2009-2010}{2008}+\frac{x-2008-2009-2010}{2009}+\frac{x-2008-2009-2010}{2010}=0\)
\(\Leftrightarrow\left(x-2008-2009-2010\right)\left(\frac{1}{2008}-\frac{1}{2009}-\frac{1}{2010}\right)=0\)
\(\Leftrightarrow x-6027=0\Leftrightarrow x=6027\)
3) Q=(3+1)(3^2+1)(3^4+1)....(3^3994+1)
=(3-1)(3+1)(3^2+1)(3^4+1)...(3^3994+1)
=(3^2-1)(3^2+1)(3^4+1)...(3^3994+1)
=(3^4-1)(3^4+1)...(3^3994+1)
=.........
=(3^3994-1)(3^3994+1)
=3^7988-1
Ta có :
\(\frac{x+1}{2012}+\frac{x+2}{2011}+\frac{x+3}{2010}=\frac{x+4}{2009}+\frac{x+5}{2008}+\frac{x+6}{2007}\)
\(\left(\frac{x+1}{2012}+1\right)+\left(\frac{x+2}{2011}+1\right)+\left(\frac{x+3}{2010}+1\right)=\left(\frac{x+4}{2009}+1\right)+\left(\frac{x+5}{2008}+1\right)+\left(\frac{x+6}{2007}+1\right)\)
\(\Leftrightarrow\)\(\frac{x+2013}{2012}+\frac{x+2013}{2011}+\frac{x+2013}{2010}=\frac{x+2013}{2009}+\frac{x+2013}{2008}+\frac{x+2013}{2007}\)
\(\Leftrightarrow\)\(\left(x+2013\right).\left(\frac{1}{2012}+\frac{1}{2011}+\frac{1}{2010}\right)=\left(x+2013\right).\left(\frac{1}{2009}+\frac{1}{2008}+\frac{1}{2007}\right)\)
\(\Leftrightarrow\)\(\frac{1}{2012}+\frac{1}{2011}+\frac{1}{2010}=\frac{1}{2009}+\frac{1}{2008}+\frac{1}{2007}\)\(\left(1\right)\)
Mà \(\frac{1}{2012}< \frac{1}{2009}\)\(;\)\(\frac{1}{2011}< \frac{1}{2008}\)\(;\)\(\frac{1}{2010}< \frac{1}{2007}\)
\(\Rightarrow\)\(\frac{1}{2012}+\frac{1}{2011}+\frac{1}{2010}< \frac{1}{2009}+\frac{1}{2008}+\frac{1}{2007}\)\(\left(2\right)\)
Từ \(\left(1\right)\)và \(\left(2\right)\)suy ra không có giá trị nào của \(x\)thoả mãn đề bài
Vậy không có gía trị nào của \(x\)hay \(x\in\left\{\varnothing\right\}\)
\(\frac{x+1}{2010}+\frac{x+2}{2009}+\frac{x+3}{2008}+...+\frac{x+2010}{1}=\left(-2010\right)\)
\(\Rightarrow\left(\frac{x+1}{2010}+1\right)+\left(\frac{x+2}{2009}+1\right)+...+\left(\frac{x+2010}{1}+1\right)=-2010+2010\)
\(\Rightarrow\frac{x+2011}{2010}+\frac{x+2011}{2009}+...+\frac{x+2011}{1}=0\)
\(\Rightarrow\left(x+2011\right)\left(1+\frac{1}{2}+...+\frac{1}{2009}+\frac{1}{2010}\right)=0\)
\(\Rightarrow x+2011=0\Leftrightarrow x=-2011\)
\(pt\Leftrightarrow\frac{x}{2009}+\frac{1}{2009}+\frac{x}{2008}+\frac{2}{2008}=\frac{x}{3}+\frac{2007}{3}+\frac{x}{4}+\frac{2006}{4}\Leftrightarrow\frac{x}{2009}+\frac{x}{2008}-\frac{x}{3}-\frac{x}{4}=\frac{2006}{4}+\frac{2007}{3}-\frac{1}{1008}-\frac{1}{2009}\Leftrightarrow x\left(\frac{1}{2009}+\frac{1}{2008}-\frac{1}{3}-\frac{1}{4}\right)=\frac{2006}{4}+\frac{2007}{3}-\frac{1}{1008}-\frac{1}{2009}\Leftrightarrow x=\frac{\frac{2006}{4}+\frac{2007}{3}-\frac{1}{1008}-\frac{1}{2009}}{\frac{1}{2009}+\frac{1}{2008}-\frac{1}{3}-\frac{1}{4}}=-2010\)
(x+1)/2011+1+(x+2)/2010+1+(x+3)/2009+1-((x+4)/2008+1+(x+5)/2007+1+(x+6)/2006+1)=0
(x+2012)/2011+(x+2012)/2010+(x+2012/2009-(x+2012)/2008-(x+2012)/2007-(x+2012)/2006=0
(x+2012)(1/2011+1/2010+1/2009-1/2008-1/2007-1/2006)=0
x+2012=0
x=-2012
lấy mỗi phân số trừ đi 1 ta đc (x-1)/2009 -1 + (x-2)/2008 -1 = (x-3)/2007-1 + (x-4)/2006 -1
suy ra (x-2010)/2009 + (x-2010)/2008 - (x-2010)/2007 - (x-2010)/2006 = 0
đặt (x-2010) làm nhân tử chung ta được
(x-2010).(1/2009 + 1/2008 - 1/2007 - 1/2006 ) = 0
=> x-2010 = 0
=> x = 2010
Dễ thấy: \(2008^3+1>0\); \(2008^2-2007>0\)
Nên \(\frac{2008^3+1}{2008^2-2007}>0\Leftrightarrow A>0\)
và \(2009-2010< 0\); \(2009^3-1>0\)
\(\Rightarrow\frac{2009^3-1}{2009-2010}< 0\Leftrightarrow B< 0\)
Vậy A > B