Đáp án:
Giải thích các bước giải:
$a,(x+3)^3-x(3x+1)^2+(2x+1)(4x^2-2x+1)=28$
$⇔x^3+9x^2+27x+27-x(9x^2+6x+1)+(8x^3+1)=28$
$⇔x^3+9x^2+27x-9x^3-6x^2-x+8x^3+1=28-27$
$⇔(x^3-9x^3+8x^3)+(9x^2-6x^2)+(27x-x)=1-1$
$⇔3x^2+26x=0$
$⇔x(3x+26)=0$
⇔\(\left[ \begin{array}{l}x=0\\3x+26=0\end{array} \right.\)
⇔\(\left[ \begin{array}{l}x=0\\x=-\dfrac{26}{3}\end{array} \right.\)
$b,(x^2-1)^3-(x^4+x^2+1)(x^2-1)=0$
$⇔x^6-3x^4+3x^2-1-(x^6-1)=0$
$⇔x^6-3x^4+3x^2-1-x^6+1=0$
$⇔(x^6-x^6)-3x^4+3x^2=0$
$⇔3x^2-3x^4=0$
$⇔3x^2(1-x^2)=0$
⇔\(\left[ \begin{array}{l}3x^2=0\\1-x^2=0\end{array} \right.\)
⇔\(\left[ \begin{array}{l}x=0\\x=±1\end{array} \right.\)
