Đáp án:
$\\\text{$x=-\dfrac{1}{2}$}; x=\dfrac{1±\sqrt{5}}{2}$
Giải thích các bước giải:
$\\\text{$$$2x^3-x^2+\sqrt[3]{2x^3-3x+1}=3x+1+\sqrt[3]{x^2+2}$$$}$ $\\\text{$$$⇔2x^3-x^2-3x-1+\sqrt[3]{2x^3-3x+1}-\sqrt[3]{x^2+2}$$$}$ $\text{Đặt $$$\sqrt[3]{2x^3-3x+1}=a⇒a^3=2x^3-3x+1$$$}$ $\\\sqrt[3]{x^2+2}=b⇒b^3=x^2+2$ $⇒a^3-b^3=2x^3-x^2-3x-1$ $\\\text{Phương tình đã cho trở thành:}$ $\text{$a^3-b^3+a-b=0$}$ $\\\text{$⇔(a-b)(a^2+ab+b^2+1)=0$}$ $\text{$⇔a-b=0 (a^2+ab+^2+1>0)$}$ $\\\text{$⇔a=b$}$ $\text{Khi $a=b$ thì $2x^3-3x+1=x^2+2$ }$ $\\\text{$⇔2x^3-x^2-3x-1=0$}$ $\text{⇔$(2x+1)(x^2-x-1)=0$}$ $\\\text{$⇔x=-\dfrac{1}{2}$}; x=\dfrac{1±\sqrt{5}}{2}$
