$x^2+x-2+\sqrt2$
Theo Vi-ét có:
$\left \{ {{x_1+x_2=-1} \atop {x_1.x_2=-2+\sqrt2}} \right.$
$A=\frac{1}{x_1} + \frac{1}{x_2} = \frac{x_1+x_2}{x_1.x_2} = \frac{-1}{-2+\sqrt2} = \frac{2+\sqrt2}{2}$
$B = |x_1-x_2| $
$⇔B^2=(x_1-x_2)^2$
$⇔B^2=x_1^2+x_2^2-2x_1.x_2$
$⇔B^2=x_1^2+x_2^2+2x_1.x_2-4x_1.x_2$
$⇔B^2=(x_1+x_2)^2-4x_1.x_2$
$⇔B^2=(-1)^2-4(-2+\sqrt2)$
$⇔B^2=9-4\sqrt2$
$⇔B=\sqrt{9-4\sqrt2}$