Cách giải:
$\dfrac{x^2}{x^2-x+1}=3x^2+4x$
$ĐK:x \neq 1$
$\to x(\dfrac{x}{(x-1)^2})=x(3x+4)$
$\to \left[ \begin{array}{l}x=0\\\dfrac{x}{(x-1)^2}=3x+4\end{array} \right.$
$\to x=(3x+4)(x-1)^2$
$\to x=(3x+4)(x^2-2x+1)$
$\to x=3x^3-6x^2+3x+4x^2-8x+4$
$\to x=3x^3-2x^2-5x+4$
$\to 3x^3-2x^2-6x+4=0$
$\to 3x^3-2x^2-2(3x-2)=0$
$\to x^2(3x-2)-2(3x-2)=0$
$\to (3x-2)(x^2-2)=0$
$\to \left[ \begin{array}{l}3x=2\\x^2=2\end{array} \right.$
$\to \left[ \begin{array}{l}x=\dfrac{2}{3}(TM)\\x=\sqrt{2}(TM)\\x=-\sqrt{2}(TM)\end{array} \right.$
