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

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

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

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

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

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 \(A_{min}=0\Leftrightarrow x=2\)

\(A=|x-1|+|x-2|+|x-3|=\left(|x-1|+|3-x|\right)+|x-2|\) \(\ge|x-1+3-x|+|x-2|\)

\(A\ge2+|x-2|\)

Vì \(|x-2|\ge0\)với \(\forall\)x

\(\Rightarrow A\ge2+0\)

\(\Rightarrow A\ge2\)

Dấu "=" xảy ra \(\Leftrightarrow\hept{\begin{cases}\left(x-1\right)\left(3-x\right)\ge0\left(1\right)\\|x-2|=0\Rightarrow x=2\left(2\right)\end{cases}}\)

Giải (1) ta có :

\(\hept{\begin{cases}x-1\ge0\\3-x\ge0\end{cases}}\Leftrightarrow\hept{\begin{cases}x\ge1\\3\ge x\end{cases}}\Leftrightarrow1\le x\le3\Leftrightarrow x\in\left\{1;2;3\right\}\left(3\right)\)

\(\hept{\begin{cases}x-1\le0\\3-x\le0\end{cases}}\Leftrightarrow\hept{\begin{cases}x\le1\\3\le x\end{cases}}\)( không có giá trị thỏa mãn )

Từ (2) và (3) => x = 2

Vậy Min_A = 2 khi x = 2

GTNN A=3

x=2; gtnn = 2

Ta có : 

\(A=\left(x-1\right)^4+\left(x-3\right)^4+6\left(x-1\right)^2\left(x-3\right)^2\)

\(A=\left(x-1\right)^4+2\left(x-1\right)^2\left(x-3\right)^2+\left(x-3\right)^4+4\left(x-1\right)^2\left(x-3\right)^2\)

\(A=\left[\left(x-1\right)^2+\left(x-3\right)^2\right]^2+4\left(x-1\right)^2\left(x-3\right)^2\)

\(A=\left[2x^2-8x+10\right]^2+4\left(x^2-4x+3\right)^2\)

\(A=\left[2\left(x-2\right)^2+2\right]+4\left[\left(x-2\right)^2-1\right]^2\)

\(A=4\left(x-2\right)^4+8\left(x-2\right)^2+4+4\left(x-2\right)^4-8\left(x-2\right)^2+4\)

\(A=8\left(x-2\right)^4+8\ge8\)

Vậy GTNN của biểu thức A là 8 \(\Leftrightarrow x=2\)

Đặt x-2=y

=> \(A=\left(y+1\right)^4+\left(y-1\right)^4+6\left(y+1\right)^2\left(y-1\right)^2\)

Khai triển A ta được 

\(A=2y^4+12y^2+2+6\left(y^4-2y^2+1\right)\)

\(=8y^4+8=8\left(y^4+1\right)\ge8\)

Dấu "=" xảy ra khi y=0 lúc đó x=0+2=2

Vậy A_min=8 khi x=2