Sagot :
A = (2x-3)²/[x(x-2)]
1) Calculer A pour
a) x=1 : (2(1)-3)²/[(1)((1)-2)] = (-1)²/-1 = -1
b) x= -3 : (2(-3)-3)²/[(-3)((-3)-2)] = (-9)²/15 = 81/15 = 27/5
c) x= 2/3: (2(2/3)-3)²/[(2/3)((2/3)-2)] = (5/3-9/3)²/[(2/3)(2/3-6/3)]
= (-4/3)²/[(2/3)(-4/3)]
= (16/9)/(-8/9)
= (16/9) * 9/-8
= -16/8
=- 2
2) Peut on calculer A pour x = 2 ?
non un diviseur ne peut pas être nul.