\(x^4-10x^2+9=0\\ \Leftrightarrow x^4-x^2-9x^2+9=0\\ \Leftrightarrow\left(x^4-x^2\right)-\left(9x^2-9\right)=0\\ \Leftrightarrow x^2\left(x^2-1\right)-9\left(x^2-1\right)=0\\ \Leftrightarrow\left(x^2-1\right)\left(x^2-9\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x^2-1=0\\x^2-9=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x^2=1\\x^2=9\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\pm1\\x=\pm3\end{matrix}\right.\\ \)
Vậy \(x=\pm1\) hoặc \(x=\pm3\)
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