\(x^4+9=5x\left(x^2-3\right)\\ \Leftrightarrow x^4+9=5x^3-15x\\ \Leftrightarrow x^4-5x^3+15x+9=0\\ \Leftrightarrow\left(x^4-3x^3\right)-\left(2x^3-6x^2\right)-\left(6x^2-18x\right)-\left(3x-9\right)=0\\ \Leftrightarrow x^3\left(x-3\right)-2x^2\left(x-3\right)-6x\left(x-3\right)-3\left(x-3\right)=0\\ \Leftrightarrow\left(x^3-2x^2-6x-3\right)\left(x-3\right)=0\\ \Leftrightarrow\left[\left(x^3+x^2\right)-\left(3x^2+3x\right)-\left(3x+3\right)\right]\left(x-3\right)=0\)
\(\Leftrightarrow\left[x^2\left(x+1\right)-3x\left(x+1\right)-3\left(x+1\right)\right]\left(x-3\right)=0\\ \Leftrightarrow\left(x^2-3x+3\right)\left(x+1\right)\left(x-3\right)=0\\ \Leftrightarrow\left[\left(x^2-3x+\dfrac{9}{4}\right)+\dfrac{3}{4}\right]\left(x+1\right)\left(x-3\right)=0\\ \Leftrightarrow\left[\left(x-\dfrac{3}{2}\right)^2+\dfrac{3}{4}\right]\left(x+1\right)\left(x-3\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}\left(x-\dfrac{3}{2}\right)^2=-\dfrac{3}{4}\left(vô.lí\right)\\x=-1\\x=3\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=-1\\x=3\end{matrix}\right.\)
