Question

помогите с заданием пожалуйста

Answer 1

$\frac{2(x^4+4x^2-12)+x^4+11x^2+30}{x^2+6}$
$\frac{2x^4+8x^2-24+x^4+11x^2+30}{x^2+6} = \frac{3x^4+19x^2+6}{x^2+6} = \frac{3x^4+18x^2+x^2+6}{x^2+6} = \\ \frac{3x^2*(x^2+6)+x^2+6}{x^2+6} = \frac{(3x^2+1)*(x^2+6)}{x^2+6} =3x^2+1$

Answer 2

(2х^4+8х²-24+х^4+11^2+30)/(x²+6)=(3x^4+19x^2+6)/(x²+6)=
=(3x²+1)(x²+6)/(x²+6)=3x²+1
3x^4+19x^2+6=0
x²=a
3a²+19a+6=3(x²+1/3)(x²+6)=(3x²+1)(x²=6)
D=361-72=289
a1=(-19-17)/6=-6⇒x²=-6
a2=(-19+17)/6=1/3