Goi A la diem thoa man dkien de bai. Vay ta co A(a, a-1).
Ta co
$y' = 3x^2 - 6x$
y'=0 <-> x = 0 hoac x=2.
Vay 2 diem cuc tri la B(0,2) va C(2,-2)
SUy ra $\vec{BA} = (a,a-3)$ va $\vec{CA} = (a-2, a+1)$.
Theo de bai ta co AB = AC hay $\vec{BA}^2 = \vec{CA^2}$
Vay
$$a^2 + (a-3)^2 = (a-2)^2 + (a+1)^2$$
<-> a = 1.
Vay diem A(1,0)
Dap an la C.