Đáp án:
Giải thích các bước giải:
$(x+3)^3-x(3x+1)^2+2x(2x-1)^2=5(x+1)(1-x)$
$⇔x^3+9x^2+27x+27-x(9x^2+6x+1)+2x(4x^2-4x+1)=-5(x-1)(x+1)$
$⇔x^3+9x^2+27x+27-9x^2-6x^2-x+8x^3-8x^2+2x=-5(x^2-1)$
$⇔(x^3+8x^3-9x^3)+(9x^2-6x^2-8x^2)+(27x-x+2x)+27=-5x^2+5$
$⇔-5x^2+5x^2+28x=5-27$
$⇔28x=-22$
$⇔x=\dfrac{-11}{14}$
$(x+1)^3-(x-1)^3-6(x-1)^2=-10$
$⇔x^3+3x^2+3x+1-x^3+3x^2-3x+1-6(x^2-2x+1)=-10$
$⇔(x^3-x^3)+(3x^2+3x^2)+(3x-3x)+(1x+1)-6x^2+12x-6=-10$
$⇔6x^2+2-6x^2+12x-6=-10$
$⇔(6x^2-6x^2)+12x=-10+6-2$
$⇔12x=-6$
$⇔x=\dfrac{-1}{2}$
