a, ĐKXĐ: $x \neq 0$
$A=\dfrac{x^5 + x^2}{x^3 - x^2 + x }$
$A=\dfrac{(x+1).x^2.(x^2-x+1)}{x(x^2-x+1)}$
$A=x(x+1)=x^2+x$
b, $A-|A|=0⇔|A|≥A⇔x(x+1)≥0⇔\begin{cases}x≤-1\\x>0\end{cases}$
c, `A=x^2+x=x^2+2.x.\frac{1}{2}+(\frac{1}{2})^2-(\frac{1}{2})^2=(x+\frac{1}{2})^2-\frac{1}{4}`
Vì `(x+\frac{1}{4})^2≥0∀x`
`⇒(x+\frac{1}{2})^2-\frac{1}{4}≥-\frac{1}{4}∀x`
Vậy: `A_{min}=-\frac{1}{4}⇔(x+\frac{1}{2})^2=0⇔x=-\frac{1}{2}`