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Bài 1: Ta có:
\(\dfrac{a+2009}{a-2009}=\dfrac{b+2010}{b-2010}\Rightarrow\left(a+2009\right)\left(b-2010\right)=\left(a-2009\right)\left(b+2010\right)\\ \Leftrightarrow ab-2010a+2009b-4038090=ab+2010a-2009b-4038090\\ \Leftrightarrow-2010a+2009b=2010a-2009b\\ \Leftrightarrow4018b=4020a\\ \Leftrightarrow1009b=1010a\\ \Leftrightarrow\dfrac{a}{2009}=\dfrac{b}{2010}\left(dpcm\right)\)
Bỏ dấu giá trị tuyệt đối:
| x \(\le\) 2008 | 2008 < x < 2009 | 2009 \(\le\) x < 2010 | 2010\(\le\)x < 2011 | x \(\ge\) 2011 | |
| |x- 2008| | 2008-x | x-2008 | x-2008 | x-2008 | x-2008 |
| |x-2009| | 2009-x | 2009-x | x-2009 | x-2009 | x-2009 |
| |x-2010| | 2010-x | 2010 - x | 2010 - x | x - 2010 | x - 2010 |
| |x-2011| | 2011 - x | 2011 - x | 2011 - x | 2011 - x | x - 2001 |
=>
+) Nếu x \(\le\) 2008 => A = 2008 - x + 2009 - x + 2010 - x + 2011 - x + 2008 = 10 046 - 4x \(\ge\) 10 046 - 4.2008 = 2014
+) Nếu 2008 < x < 2009 => A = x - 2008 + 2009 - x + 2010 - x + 2011 - x + 2008 = 6030 - 2x > 6030 - 2.2009 = 2012
+) Nếu 2009 \(\le\) x < 2010 => A = x - 2008 + x - 2009 + 2010 - x + 2011 - x + 2008 = 2012
+) Nếu 2010 \(\le\) x < 2011 => A = x - 2008 + x - 2009 + x - 2010 + 2011 - x + 2008 = 2x - 2008 \(\ge\) 2.2010 - 2008 = 2012
+) Nếu x \(\ge\) 2011 => A = x - 2008 + x - 2009 + x - 2010 + x - 2011 + 2008 = 4x - 6030 \(\ge\) 4.2011 - 6030 = 2014
Từ các trường hợp trên => A nhỏ nhất bằng 2012 khi x = 2009 ; hoặc x = 2010
\(\frac{x-1}{2011}+\frac{x-2}{2010}-\frac{x-3}{2009}=\frac{x-4}{2008}\)
\(\Rightarrow\left(\frac{x-1}{2011}+1\right)+\left(\frac{x-2}{2010}+1\right)-\left(\frac{x-3}{2009}+1\right)=\frac{x-4}{2008}+1\)
\(\Rightarrow\frac{x-1+2011}{2011}+\frac{x-2+2010}{2010}-\frac{x-3+2009}{2009}=\frac{x-4+2008}{2008}\)
\(\Rightarrow\frac{x-2012}{2011}+\frac{x-2012}{2010}-\frac{x-2012}{2009}=\frac{x-2012}{2008}\)
\(\Rightarrow\frac{x-2012}{2011}+\frac{x-2012}{2010}-\frac{x-2012}{2009}-\frac{x-2012}{2008}=0\)
\(\Rightarrow\left(x-2012\right)\left(\frac{1}{2011}+\frac{1}{2010}-\frac{1}{2009}-\frac{1}{2008}\right)=0\)
Mà \(\frac{1}{2011}+\frac{1}{2010}-\frac{1}{2009}-\frac{1}{2008}\ne0\)
=> x - 2012 = 0
=> x = 2012
Vậy x = 2012
\(\frac{x-1}{2011}+\frac{x-2}{2010}-\frac{x-3}{2009}=\frac{x-4}{2008}\)
\(\Rightarrow\left(\frac{x-1}{2011}+1\right)+\left(\frac{x-2}{2010}+1\right)-\left(\frac{x-3}{2009}+1\right)=\frac{x-4}{2008}+1\)
\(\Rightarrow\frac{x-1+2011}{2011}+\frac{x-2+2010}{2010}-\frac{x-3+2009}{2009}=\frac{x-4+2008}{2008}\)
\(\Rightarrow\frac{x-2012}{2011}+\frac{x-2012}{2010}-\frac{x-2012}{2009}=\frac{x-2012}{2008}\)
\(\Rightarrow\frac{x-2012}{2011}+\frac{x-2012}{2010}-\frac{x-2012}{2009}-\frac{x-2012}{2008}=0\)
\(\Rightarrow\left(x-2012\right)\left(\frac{1}{2011}+\frac{1}{2010}-\frac{1}{2009}-\frac{1}{2008}\right)=0\)
Mà \(\frac{1}{2011}+\frac{1}{2010}-\frac{1}{2009}-\frac{1}{2008}\ne0\)
=> x - 2012 = 0
=> x = 2012
Vậy x = 2012
Kết quả đúng òi nhưng mà dấu suy ra thứ 2 ế \(x-1+2011\) thì bằng \(x+2010\) mà. Cả mấy cái bên cạnh cũng bị tính sai. 
\(\frac{x+10}{2008}+\frac{x+9}{2009}=\frac{x+8}{2010}+\frac{x+7}{2011}\)
\(\left(\frac{x+10}{2008}+1\right)+\left(\frac{x+9}{2009}+1\right)=\left(\frac{x+8}{2010}+1\right)+\left(\frac{x+7}{2011}+1\right)\)
\(\frac{x+2018}{2008}+\frac{x+2018}{2009}=\frac{x+2018}{2010}+\frac{x+2018}{2011}\)
\(\frac{x+2018}{2008}+\frac{x+2018}{2009}-\frac{x+2018}{2010}-\frac{x+2018}{2011}=0\)
\(x+2018\cdot\left(\frac{1}{2008}+\frac{1}{2009}-\frac{1}{2010}-\frac{1}{2011}\right)=0\)
mà \(\left(\frac{1}{2008}+\frac{1}{2009}-\frac{1}{2010}-\frac{1}{2011}\right)\ne0\)
\(\Rightarrow x+2018=0\)
\(\Rightarrow x=-2018\)
Vậy,.............
Ta có: \(\frac{x+10}{2008}+\frac{x+9}{2009}=\frac{x+8}{2010}+\frac{x+7}{2011}\)
\(\Rightarrow\frac{x+10}{2008}+1+\frac{x+9}{2009}+1=\frac{x+8}{2010}+1+\frac{x+7}{2011}+1\)
\(\Rightarrow\frac{x+2018}{2008}+\frac{x+2018}{2009}=\frac{x+2018}{2010}+\frac{x+2018}{2011}\)
\(\Rightarrow\frac{x+2018}{2008}+\frac{x+2018}{2009}-\frac{x+2018}{2010}-\frac{x+2018}{2011}=0\)
\(\Rightarrow x+2018\cdot\left(\frac{1}{2008}+\frac{1}{2009}-\frac{1}{2010}-\frac{1}{2011}\right)=0\)
Do \(\frac{1}{2008}+\frac{1}{2009}-\frac{1}{2010}-\frac{1}{2011}\ne0\)
\(\Rightarrow x+2018=0\)
\(\Rightarrow x=-2018\)
Vậy \(x=-2018\)