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Đáp án:
$a)$
$4x^2 - 12x + 11$
$= (4x^2 - 12x + 9) + 2$
$= (2x - 3)^2 + 2 ≥ 2 > 0$
$b)$
$x^2 - x + 1$
$= (x^2 - x + \dfrac{1}{4}) + \dfrac{3}{4}$
$= (x - \dfrac{1}{2})^2 + \dfrac{3}{4} ≥ \dfrac{3}{4} > 0$
$c)$
$x^2 - 2x + y^2 + 4y + 6$
$= (x^2 - 2x + 1) + (y^2 + 4y + 4) + 1$
$= (x - 1)^2 + (y + 2)^2 + 1 ≥ 1 > 0$