Question

x^4+ y^4 при х +у =6 . ху = -3

Answer 1

{x+y=6⇒x=6-y
{xy=-3
(6-y)*y=-3
y²-6y-3=0
D=36+12=48
√D=4√3
y1=(6-4√3)/2=3-2√3⇒x1=6-3+2√3=3+2√3
y2=3+2√3⇒x2=6-3-2√3=3-2√3
(x1)^4=[(3+2√3)²]²=(9+12√3+12)²=(21+12√3)²=441+504√3+432=
=873+504√3
(y1)²=[(3-2√3)²]²=873-504√3
x^4+y^4=873+504√3+873-504√3=1746