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Lấy logarit 2 vế ta có: $(x^2+2x)\ln(x^2-x+1)\le0$
$\Leftrightarrow \left[\begin{array}{l} x^2+2x=0\\\ln(x^2-x+1)=0\\\left\{ \begin{array}{l} x^2+2x>0\\\ln(x^2-x+1)<0\end{array} \right.\\\left\{ \begin{array}{l} x^2+2x<0\\\ln(x^2-x+1)>0\end{array} \right. \end{array} \right.$
$\Leftrightarrow \left[\begin{array}{l} x=0\\x=-2\\x=1\\\left\{ \begin{array}{l} x^2+2x>0\\0<x^2-x+1<1\end{array} \right.\\\left\{ \begin{array}{l} x^2+2x<0\\x^2-x+1>1\end{array} \right. \end{array} \right.$
$\Leftrightarrow \left[ \begin{array}{l} x=0\\x=-2\\x=1\\0<x<1\\-2<x<0\end{array} \right. \Leftrightarrow -2\le x\le 1$
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