Giải thích các bước giải:
a.$A=(x^2-1)(x+2)-(x-2)(x^2+2x+4)$
$\to A=x^3+2x^2-x-2-(x^3-2^3)$
$\to A=2x^2-x+6$
b.$B=x^2(x+y)+y^2(x+y)+2x^2y+2y^2x$
$\to B=x^3+x^2y+xy^2+y^3+2x^2y+2y^2x$
$\to B=x^3+3x^2y+3xy^2+y^3$
$\to B=(x+1)^3$
c.$\dfrac{x^3+4x^2-x-4}{x+1}$
$=\dfrac{\left(x^3+4x^2\right)+\left(-x-4\right)}{x+1}$
$=\dfrac{\left(x+4\right)\left(x^2-1\right)}{x+1}$
$=\dfrac{(x+4)(x+1)(x-1)}{x+1}$
$=(x+4)(x-1)$
d.$\dfrac{x^4-2x^2+2x-1}{x^2-1}$
$=\dfrac{\left(x-1\right)\left(x^3+x^2-x+1\right)}{(x-1)(x+1)}$
$=\dfrac{x^3+x^2-x+1}{x+1}$