\(x^3-5x^2+8x-4=0\Leftrightarrow x^3-x^2-4x^2+4x+4x-4=0\Leftrightarrow x^2\left(x-1\right)-4x\left(x-1\right)+4\left(x-1\right)=0\Leftrightarrow\left(x^2-4x+4\right)\left(x-1\right)=0\Leftrightarrow\left(x-2\right)^2\left(x-1\right)=0\Leftrightarrow\left[{}\begin{matrix}\left(x-2\right)^2=0\\\left(x-1\right)=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x-2=0\\x=1\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=2\left(tm\right)\\x=1\left(tm\right)\end{matrix}\right..Vậy:x\in\left\{1;2\right\}\)
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