a) $y = x^3 (1-x)^2$
$y' = 3x^2 (1-x)^2 - 2x^3 (1-x) = 5 x^4 - 8 x^3 + 3 x^2 = x^2 (5x^2 - 8x + 3)$
y'=0 <-> x=0, x=1, va x=3/5.
Do x=0 co boi la 2 nen hso ko doi dau tai x=0.
Vay cac diem cuc tri la x=1, x=3/5.
b) $y = (x+2)^2 (x-3)^3$
$y' = 2(x+2)(x-3)^3 + 3(x+2)^2(x-3)^2 = 5x (x+2) (x-3)^2$
y'=0 <-> x = 0, x = -2 va x=3.
Do x=3 co boi la 2 nen hso ko doi dau tai x=3.
Vay cac diem cuc tri la x=0, x=-2.
