Tìm GTNN : A = $\dfrac{x}{x^{2} - 2x + 1}$

$\text{A = $\dfrac{x}{x^{2} - 2x + 1}$}$
$\text{A = $\dfrac{x - 1 + 1}{(x - 1)^{2}}$}$
$\text{A = $\dfrac{x - 1}{(x - 1)^{2}}$ + $\dfrac{1}{(x - 1)^2}$}$
$\text{A = $\dfrac{1}{x - 1}$ + $\dfrac{1}{(x - 1)^{2}}$}$
$\text{Đặt a = $\dfrac{1}{x - 1}$}$
$\text{A = $a^{2}$ + a}$
$\text{A = $a^{2}$ + 2 . a . $\dfrac{1}{2}$ + $(\dfrac{1}{2})^{2}$ - $(\dfrac{1}{2})^{2}$}$
$\text{A = $(a + \dfrac{1}{2})^{2}$ - $\dfrac{1}{4}$ ≥ $\dfrac{-1}{4}$}$
$\text{Dấu "=" xảy ra khi :}$
$\text{$(a + \dfrac{1}{2})^{2}$ = 0}$
$\text{⇔ a + $\dfrac{1}{2}$ = 0}$
$\text{⇔ a = $\dfrac{-1}{2}$}$
$\text{⇔ $\dfrac{1}{x - 1}$ = $\dfrac{-1}{2}$}$
$\text{⇔ -1(x - 1) = 1 . 2}$
$\text{⇔ -x + 1 = 2}$
$\text{⇔ -x = 1}$
$\text{⇔ x = -1}$

$\text{Vậy GTNN của A là $\dfrac{-1}{4}$ khi x = -1}$