25 $x^{2}$ + 5x + 29
= $(5x)^{2}$ + 2 . 5x . $\dfrac{1}{2}$ + $(\dfrac{1}{2})^2$ - $(\dfrac{1}{2})^2$ + 29
= $(5x + \dfrac{1}{2})^2$ + 28 + $\dfrac{3}{4}$
Ta có: $(5x + \dfrac{1}{2})^2$ ≥ 0 ∀ x
⇔ $(5x + \dfrac{1}{2})^2$ + 28 + $\dfrac{3}{4}$ ≥ 28 + $\dfrac{3}{4}$ ∀ x
⇔ $(5x + \dfrac{1}{2})^2$ + 28 + $\dfrac{3}{4}$ > 0 ∀ x
⇔ 25 $x^{2}$ + 5x + 29 > 0 ∀ x
Vậy đa thức trên luôn dương vs mọi x
