\(D=\frac{3x^2-4x+8}{x^2+2}\)
\(=\frac{2x^2+4}{x^2+2}+\frac{x^2-4x+4}{x^2+2}\)
\(=\frac{2\left(x^2+2\right)}{x^2+2}+\frac{\left(x-2\right)^2}{x^2+2}\)
\(=2+\frac{\left(x-2\right)^2}{x^2+2}\ge2\)
Dấu ''='' xả ra khi x - 2 = 0
<=> x = 2
Vậy Min D = 2 khi x = 2