a,The value of A=\(2\times\left|2008-2007\right|+\left|2008-2009\right|\)\(=\)\(2\times\left|1\right|+\left|-1\right|=2\times1+1=2+1=3\)

b,With \(x< 2007\)then A=x-2010

\(\Rightarrow2\times-\left(x-2007\right)-\left(x-2009\right)=x-2010\)

\(\Leftrightarrow-2x+4014-x+2009=x-2010\)\(\Leftrightarrow-4x=-2010-4014-2009\Leftrightarrow-4x=-8033\)

\(\Leftrightarrow x=\dfrac{8033}{4}>\dfrac{8028}{4}=2007\)(not satisfy)

With \(2007\le x< 2009\) then \(2\times\left(x-2007\right)-\left(x-2009\right)=x-2010\)

\(\Leftrightarrow2x-4014-x+2009=x-2010\)

\(\Leftrightarrow2x-x-x=-2010-2009+4014\)\(\Leftrightarrow0=-5\)(not satisfy)

With \(x\ge2009\) then 

\(2\times\left(x-2007\right)+\left(x-2009\right)=x-2010\)

\(\Leftrightarrow2x-4014+x-2009=x-2010\)

\(\Leftrightarrow2x=-2010+2009+4014\Leftrightarrow2x=4013\)

\(\Leftrightarrow x=2006,5< 2009\)(not satisfy)

 So not have x satisfy

c,We have:A=\(\left|x-2007\right|+\left|x-2009\right|+\left|x-2007\right|\)

\(=\left|x-2007\right|+\left|2009-x\right|+\left|x-2007\right|\)\(\ge\left|x-2007+2009-x\right|+\left|x-2007\right|=\left|2\right|+\left|x-2007\right|\ge2\)

So the smallest of A is 2 when \(x=2007\)

a) With x = 2008 => \(A=2\cdot\left|2008-2007\right|+\left|2008-2009\right|=2+1=3\)

b) Cause A = \(2\cdot\left|x-2007\right|+\left|x-2009\right|\)

So for A = x - 2010

<=> \(2\cdot\left|x-2007\right|+\left|x-2009\right|=x-2010\)   (*)

+) With x < 2007 

(*) <=> 2(2007 - x) + (2009 - x) = x - 2010 

     <=>  6023 - 3x = x - 2010

     <=>  6023 + 2010 = 4x

     <=>  x = \(\dfrac{8033}{4}>2007\)  (Removed) 

+) With \(2007\le x\le2009\)

(*) <=> \(2\cdot\left(x-2007\right)+\left(2009-x\right)=x-2010\)

     <=> \(x-2005=x-2010\) (impossible) 

+) With \(x>2009\)

(*)  <=> \(2\cdot\left(x-2007\right)+\left(x-2009\right)=x-2010\)

      <=>  \(3x-6023=x-2010\)

      <=> \(2x=4013\)

      <=>  \(2x=\dfrac{4013}{2}< 2009\)  (Removed)

So , There isn't any roots of this equation 

c) Applying this inequality 

With a,b are two numbers , we have \(\left|a\right|+\left|b\right|\ge\left|a+b\right|\)

Equation occur <=> ab \(\ge\) 0

So \(A=2\cdot\left|x-2007\right|+\left|x-2009\right|\)

         \(=\left|x-2007\right|+\left|2009-x\right|+\left|x-2007\right|\ge\left|x-2007+2009-x\right|+\left|x-2007\right|\)

                                                                                  \(=2+\left|x-2007\right|\ge2\)

=> Min(A) = 2 <=> \(\left\{{}\begin{matrix}x=2007\\\left(x-2007\right)\left(2009-x\right)\ge0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=2007\\2007\le x\le2009\end{matrix}\right.\)

So Min(A) = 2 <=> x = 2007