\(A=\left|x-2006\right|+\left|x-1\right|=\left|x-2006\right|+\left|-x+1\right|\ge\left|x-2006-x+1\right|=2005\)

dấu = xảy ra khi \(\left(x-2006\right).\left(-x+1\right)\ge0\)

\(\Rightarrow1\le x\le2006\)

Vậy Min A=2015 khi và chỉ khi \(1\le x\le2006\)

\(A=|x-2006|+|x-1|=|x-2006|+|1-x|\)

\(\Rightarrow A\ge|x-2006+1-x|=|-2005|=2005\)

\(\Rightarrow minA=2005\Leftrightarrow\left(x-2006\right).\left(1-x\right)\ge0\)

\(TH1:\hept{\begin{cases}x-2006< 0\\1-x< 0\end{cases}}\Rightarrow\hept{\begin{cases}x< 2006\\1< x\end{cases}}\Rightarrow\hept{\begin{cases}x< 2006\\x>1\end{cases}}\Rightarrow1< x< 2006\left(t/m\right)\)

\(TH2:\hept{\begin{cases}x-2006\ge0\\1-x\ge0\end{cases}}\Rightarrow\hept{\begin{cases}x\ge2006\\1\ge x\end{cases}}\Rightarrow\hept{\begin{cases}x\ge2006\\x\le1\end{cases}}\)(vô lý) 

Vậy \(minA=2005\Leftrightarrow1< x< 2006\)

Áp dụng BĐT sau : \(\left|a\right|+\left|b\right|\ge\left|a+b\right|\)

Ta có : 

\(A=\left|x-2006\right|+\left|x-1\right|=\left|x-2006\right|+\left|1-x\right|\ge\left|x-2006+1-x\right|\)

\(\Rightarrow A\ge2005\)

Dấu''=''  xảy ra \(\Leftrightarrow\left(x-2006\right)\left(1-x\right)\ge0\)

\(\Leftrightarrow\orbr{\begin{cases}x-2006\ge0\\1-x\ge0\end{cases}}\)hoặc \(\orbr{\begin{cases}x-2006< 0\\1-x< 0\end{cases}}\)

\(\Leftrightarrow\hept{\begin{cases}x\ge2006\\x\le1\end{cases}}\)hoặc \(\hept{\begin{cases}x< 2006\\x>1\end{cases}}\)(loại ) 

\(\Leftrightarrow1\le x\le2006\)

Vậy ..................

Áp dụng BĐT \(\left|a\right|+\left|b\right|\ge\left|a+b\right|\):

\(A=\left|x-3\right|+\left|x-1\right|+\left|x+1\right|+\left|x+3\right|\)

\(=\left|3-x\right|+\left|x+3\right|+\left|1-x\right|+\left|x+1\right|\)

\(\ge\left|3-x+x+3\right|+\left|1-x+x+1\right|=8\)

\(minA=8\Leftrightarrow\left\{{}\begin{matrix}\left(3-x\right)\left(x+3\right)\ge0\\\left(1-x\right)\left(x+1\right)\ge0\end{matrix}\right.\Leftrightarrow-1\le x\le1\)

a) GTNN = 0 khi x = -1

b) GTNN = 503 khi x =0

Đặt x -2006 = y 

pt <=>  \(\frac{y^2-y\left(y-1\right)+\left(y-1\right)^2}{y^2+y\left(y-1\right)+\left(y-1\right)^2}=\frac{19}{49}\)

<=> \(\frac{y^2-y^2+y+y^2-2y+1}{y^2+y^2-y+y^2-2y+1}=\frac{19}{49}\)

<=> \(\frac{y^2-y+1}{3y^2-3y+1}=\frac{19}{49}\)

<=> \(49y^2-49y+49=57y^2-57y+19\)

<=> \(8y^2-8y-30=0\)

<=> \(4y^2-4y+15=0\)

Giải tiếp nha 

Đặt \(x+3=t\ne0\Rightarrow x=t-3\)

\(A=\dfrac{\left(t+2\right)\left(t-4\right)}{t^2}=\dfrac{t^2-2t-8}{t^2}=-\dfrac{8}{t^2}-\dfrac{2}{t}+1=-8\left(\dfrac{1}{t}+\dfrac{1}{8}\right)^2+\dfrac{9}{8}\le\dfrac{9}{8}\)

\(A_{max}=\dfrac{9}{8}\) khi \(t=-8\) hay \(x=-11\)

Ta có tính chất : 

\(\left|a\right|+\left|b\right|\ge\left|a+b\right|\)

\(\rightarrow A=\left|x+5\right|+\left|x+2\right|+\left|x-7\right|+\left|x-8\right|\ge\left|x+5+x+2+x-7+x-8\right|\)

​​\(\rightarrow A\ge\left|4x-8\right|\)

Vì \(\left|4x-8\right|\ge0\forall x\in R\) nên :

\(\rightarrow A\ge0\forall x\in R\)

Dấu "= " xảy ra khi : 

\(\left|4x-8\right|=0\) \(\Leftrightarrow4x-8=0\) 

                     \(\Leftrightarrow x=2\)

Vậy \(A_{min}=0\Leftrightarrow x=2\)