Violympic hả em???
Bài làm:
\(\left(x+1\right)\left(x-2\right)^2+x^2\left(4-x\right)=13\\ < =>\left(x+1\right)\left(x^2-4x+4\right)+x^2\left(4-x\right)=13\\ < =>x^3-4x^2+4x+x^2-4x+4+4x^2-x^3=13\\ < =>x^2=13-4=9\\ Mà:\left\{{}\begin{matrix}3^2=9\\\left(-3\right)^2=9\end{matrix}\right.< =>\left\{{}\begin{matrix}x^2=3^2\\x^2=\left(-3\right)^2\end{matrix}\right.=>\left\{{}\begin{matrix}x=3\\x=-3\end{matrix}\right.\)
Vậy đáp án: -3;3