Bạn tham khảo nhé:
⇔$\frac{x-1}{x-2}$= $\frac{-12}{(x-2)(x^2+2x+4)}$
⇔$\frac{(x-1)(x^2+2x+4)}{(x-2)(x^2+2x+4)}$= $\frac{-12}{(x-2)(x^2+2x+4)}$
⇔$\frac{x^3+2x^2+4x-x^2-2x-4}{(x-2)(x^2+2x+4)}$=$\frac{-12}{(x-2)(x^2+2x+4)}$
⇔$\frac{x^3+x^2+2x-4}{(x-2)(x^2+2x+4)}$=$\frac{-12}{(x-2)(x^2+2x+4)}$
⇒$x^{3}$+ $2x^{2}$- $x^{2}$-$2x$+$4x$+$8$=$0$ (làm tắt nhé)
⇔$x^{2}$$(x+2)$-$x(x+2)$+$2(x+2)$=$0$
⇔($x^{2}$-$x+2$)(x+2)=0
⇔\(\left[ \begin{array}{l}x^2-x+2=0(loại)\\x=-2\end{array} \right.\)