a) $6x^2-9x$
$=3x(2x-3)$
b) $x^3-2x^2+3x-6$
$=x^2(x-2)+3(x-2)$
$=(x-2)(x^2+3)$
c) $x^2-4x+4-9y^2$
$=(x^2-4x+4)-9y^2$
$=(x^2-2\cdot x\cdot2+2^2)-(3y)^2$
$=(x-2)^2-(3y)^2$
$=(x-2-3y)(x-2+3y)$
d) $3x^2-4x+1$
$=3x^2-3x-x+1$
$=3x(x-1)-(x-1)$
$=(x-1)(3x-1)$
e) $3x^3-5x^2-16x+12$
$=3x^3-9x^2+4x^2-12x-4x+12$
$=3x^2(x-3)+4x(x-3)-4(x-3)$
$=(x-3)(3x^2+4x-4)$
$=(x-3)(3x^2+6x-2x-4)$
$=(x-3)[3x(x+2)-2(x+2)]$
$=(x-3)(x+2)(3x-2)$
Mình đã trả lời câu này rồi nhé!
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