Bonjour,
f existe ssi x²+x+1 ≠ 0
'
Résoudre f(x)= 2
(3x²-2)/ (x²+x+1)= 2
3x²-2= 2 (x²+x+1)
3x²-2- 2 (x²+x+1)= 0
3x²-2- 2 x²-2x-2= 0
x²-2x-4= 0
Δ= b²-4ac= (-2)²-4(1)(-4)= 4+16= 20 > 0 ; 2 solutions ***√(4x5)= 2√5
x1= (-(-2)-√20)/2(1)= (2-2√5)/2= 1-√5
x2= (2+2√5)/2= 1+√5
S= { 1-√5; 1+√5 }
