\(P=\frac{12x^2-6x+4}{x^2+1}=\frac{\left(9x^2-6x+1\right)+3\left(x^2+1\right)}{x^2+1}=\frac{\left(3x-1\right)^2}{x^2+1}+3\ge3\forall x\)

Dấu "=" xảy ra khi: \(3x-1=0\Rightarrow x=\frac{1}{3}\)

Vậy \(P_{min}=3\Leftrightarrow x=\frac{1}{3}\)

\(A=x^2-2x+4\)

\(A=x^2-2x+1+3=\left(x-1\right)^2+3\ge3\)

dấu = xảy ra khi x-1=0

=> x=1

Vậy MinA=3 khi x=1

Ta có :

A = x⁴ - 2x² + x² + 2x + 1 + 2019

A = ( x² - 1 )² + ( x + 1 )² + 2019 \(\ge\)2019

Vậy GTNN của A là 2019 \(\Leftrightarrow\hept{\begin{cases}x+1=0\\x^2-1=0\end{cases}\Leftrightarrow x=-1}\)

\(\frac{2014a}{ab+2014a+2014}+\frac{b}{bc+b+2014}+\frac{c}{ac+c+1}\)

\(=\frac{abc.a}{ab+abca+abc}+\frac{b}{bc+b+abc}+\frac{c}{ac+c+1}\)

\(=\frac{ac}{1+ac+c}+\frac{1}{c+1+ac}+\frac{c}{ac+c+1}=1\left(ĐPCM\right)\)

Ta có \(\frac{x^2+1}{x^3-1}=\frac{a}{x-1}+\frac{bx+c}{x^2+x+1}\)

\(\Rightarrow\frac{x^2+1}{x^3-1}=\frac{a.\left(x^2+x+1\right)}{x^3-1}+\frac{\left(x-1\right).\left(bx+c\right)}{x^3-1}\)

\(\Rightarrow\frac{x^2+1}{x^3-1}=\frac{ax^2+ax+a}{x^3-1}+\frac{bx^2-xc-xb-c}{x^3-1}\)

\(\Rightarrow\frac{x^2+1}{x^3-1}=\frac{x^2.\left(a+b\right)+x.\left(a-b-c\right)+\left(a-c\right)}{x^3-1}\)

Đồng nhất hệ số hai vế của tử số ta có 

\(\hept{\begin{cases}a+b=1\\a-b-c=0\\a-c=1\end{cases}\Rightarrow}\hept{\begin{cases}a+b=1\\a-c=b\\a-c=1\end{cases}\Rightarrow\hept{\begin{cases}a=1\\b=1\\c=0\end{cases}}}\)