Giải hệ phương trình bậc 3 online

$a{x}^3+b{x}^2+cx+d=0\left(a<>0\right)$

x³ +

x² +

x +

=

$2x^3+5x^2+\mathrm{(-}\mathrm{1)}x+\mathrm{(-}\mathrm{6)}=0$

$*\; △=b^2-3ac$

$=5^2-3.2.-1$

$=31$

$*k=\frac{9abc-2b^3-27a^{2}d}{2\sqrt{|△{|}^3}}$

$=\frac{-90- 250-(-648)}{2\sqrt{|31{|}^3}}$

$=\mathrm{|0,89|}$

$=>$
$x_{1=}\frac{2\sqrt{31}\cos ({\displaystyle \frac{arc\cos (\mathrm{0,89})}{3}})-b}{3a}=1$
$x_{2=}\frac{2\sqrt{31}\cos ({\displaystyle \frac{arc\cos (\mathrm{0,89})}{3}}-{\displaystyle \frac{2\mathrm{\pi }}{3}})-b}{3a}=\mathrm{-1,5}$
$x_{3=}\frac{2\sqrt{31}\cos ({\displaystyle \frac{arc\cos (\mathrm{0,89})}{3}}+{\displaystyle \frac{2\mathrm{\pi }}{3}})-b}{3a}=\mathrm{-2}$