Giải thích các bước giải:
$b)2x(x+2)^2-8x^2=2(x-2)(x^2+2x+4)$
$⇒2x(x^2+4x+4)-8x^2=2(x^3-8)$
$⇒2x^3+8x^2+8x-8x^2=2x^3-16$
$⇒8x=-16$
$⇒x=-2$
Vậy $x=-2$
$d)(x-2)^3+(3x-1)(3x+1)=(x+1)^3$
$⇒x^3-6x^2+12x-8+9x^2-1=x^3+3x^2+3x+1$
$⇒12x-9=3x+1$
$⇒9x=10$
$⇒x=\dfrac{10}{9}$
Vậy $x=\dfrac{10}{9}$
$f)(x-1)^3-x(x+1)^2=5x(2-x)-11(x+2)$
$⇒x^3-3x^2+3x-1-x(x^2+2x+1)=10x-5x^2-11x-22$
$⇒x^3-3x^2+3x-1-x^3-2x^2-x=-5x^2-x-22$
$⇒2x-1=-x-22$
$⇒3x=-21$
$⇒x=-7$
Vậy $x=-7$
$h)(x-3)(x+4)-2(3x-2)=(x-4)^2$
$⇒x^2+x-12-6x+4=x^2-8x+16$
$⇒-5x-8=-8x+16$
$⇒3x=24$
$⇒x=8$
Vậy $x=8$
$j)(x+1)(x^2-x+1)-2x=x(x+1)(x-1)$
$⇒x^3+1-2x=x(x^2-1)$
$⇒x^3-2x+1=x^3-x$
$⇒x=1$
Vậy $x=1$
