Exercice 1 :
1)
(z1)^2 = (1/2)^2 x (-1 + iV3)^2
(z1)^2 = 1/4 x (1 - i 2V3 - 3)
(z1)^2 = 1/4 x (-2 - i 2V3)
(z1)^2 = 1/2 x (-1 - iV3)
(z1)^2 = z2
2)
(z2)^2 = (1/2)^2 x (-1 - iV3)^2
(z2)^2 = 1/4 x (1 + i 2V3 - 3)
(z2)^2 = 1/4 x (-2 + i 2V3)
(z2)^2 = 1/2 x (-1 + iV3)
(z2)^2 = z1
3) z1 + z2 + 1
1/2 x (-1 + iV3) + 1/2 x (-1 - iV3) + 1
(-1/2) + (iV3)/2 - 1/2 - (iV3)/2 + 1
(-1) + 1 = 0
4) je te laisse le faire
5) distances AB, AC et BC :
AB = |z2 - z1|
AB = |1/2 x (-1 - iV3) - 1/2 x (-1 + iV3)|
AB = |-1/2 - (iV3)/2 + 1/2 - (iV3)/2|
AB = |-iV3|
AB = V(V3)^2
AB = V3
AC = |1 - 1/2 x (-1 + iV3)|
AC = |1 + 1/2 - (iV3)/2|
AC = |3/2 - (iV3)/2|
AC = V[(3/2)^2 - (V3/2)^2]
AC = V(9/4 - 3/4)
AC = V(5/4)
AC = 1/2 V5
Je te laisse calculer le dernier
Exercice 2 :
f(x) = -2x^2 + 5x - 3 = 0
On calcule le discriminant :
Delta = (5)^2 - 4 * (-2) * (-3)
Delta = 25 - 24
Delta = 1 > 0 donc deux solutions
X1 = (-5 - V1) / (2 * (-2))
X1 = (-5 - 1) / (-4)
X1 = (-6) / (-4)
X1 = (3/2)
X2 = (-5 + V1) / (2 * (-2))
X2 = (-5 + 1) / (-4)
X2 = (-4) / (-4)
X2 = 1
2)
x .........|...-inf.............1.................3/2...........+inf
-------------------------------------------------------------
f(x)......|............(-)......O.....(+)........O.....(-).............
3) f(x) >= 0 :
x € [ 1 ; 3/2 ]
