\(A=\dfrac{27-12x}{x^2+9}=\dfrac{x^2+9+27-12x}{x^2+9}-1=\dfrac{x^2-12x+36}{x^2+9}-1=\dfrac{\left(x-6\right)^2}{x^2+9}-1\ge-1\)
Dấu = xảy ra khi x = 6
Vậy:...
A= \(\dfrac{27-12x}{x^2-9}\)
= \(\dfrac{x^2-12x+36-x^2-9}{x^2-9}\)
= \(\dfrac{\left(x-6\right)^2-\left(x^2+9\right)}{x^2+9}\)
= \(\dfrac{\left(x-6\right)^2}{x^2+9}-1\)
Ta có \(\dfrac{\left(x-6\right)^2}{x^2+9}\) ≥ 0 ∀ x
⇒ \(\dfrac{\left(x-6\right)^2}{x^2+9}-1\) ≥ -1 ∀ x
Vậy AMin= -1 tại x=6