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

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 \(x=-11\)

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

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

Đặt \(x^2+x=k\)

Lúc đó \(A=k\left(k-4\right)\)

\(=k^2-4k+4-4=\left(k-2\right)^2-4\ge-4\)

(Dấu "=" xảy ra khi \(k=2\Leftrightarrow x^2+x=2\)

\(\Leftrightarrow x^2+x-2=0\)

Ta có: \(\Delta=1^2+4.2=9,\sqrt{\Delta}=3\)

\(\Rightarrow\orbr{\begin{cases}x=\frac{-1+3}{2}=1\\x=\frac{-1-3}{2}=-2\end{cases}}\))

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

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

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

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

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

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 \(A_{min}=0\Leftrightarrow x=2\)

≥ 0 ∀ x

≥ 0 ∀ x

6(x−1)²(x−2)² ≥ 0 ∀ x

=> ≥ 0 ∀ x

=>A≥ 0 ∀ x

=>A_min=0. Dấu "=" xảy ra khi và chỉ khi :

⇔x=1 và (x−3)⁴=0 ⇔ x=3 và 6(x−1)²(x−2)²⇔ x=1 hoặc x=2

Vì x chỉ có 1 giá trị duy nhất trong biểu thức nên x = ∅.

Đặt \(x-2=a\Rightarrow\left\{{}\begin{matrix}x-1=a+1\\x-3=a-1\end{matrix}\right.\)

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

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

\(A=2a^4+12a^2+6a^2\left(a-1\right)^2+2\ge2\)

\(\Rightarrow A_{min}=2\) khi \(a=0\Leftrightarrow x=2\)