\(x^4+9=5x\left(3-x^2\right)\)
\(\Leftrightarrow x^4+9=15x-5x^3\)
\(\Leftrightarrow x^4+5x^3-15x+9=0\)
\(\Leftrightarrow x^4-x^3+6x^3-6x^2+6x^2-6x-9x+9=0\)
\(\Leftrightarrow\left(x^4-x^3\right)+\left(6x^3-6x^2\right)+\left(6x^2-6x\right)-\left(9x-9\right)=0\)
\(\Leftrightarrow x^3\left(x-1\right)+6x^2\left(x-1\right)+6x\left(x-1\right)-9\left(x-1\right)=0\)
\(\Leftrightarrow\left(x-1\right)\left(x^3+6x^2+6x-9\right)=0\)
\(\Leftrightarrow\left(x-1\right)\left(x^3+3x^2+3x^2+9x-3x-9\right)=0\)
\(\Leftrightarrow\left(x-1\right)\left[x^2\left(x+3\right)+3x\left(x+3\right)-3\left(x+3\right)\right]=0\)
\(\Leftrightarrow\left(x-1\right)\left(x+3\right)\left(x^2+3x-3\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x-1=0\\x+3=0\\x^2+3x-3=0\end{matrix}\right.\)
Ta có: \(x^2+3x-3=0\)
\(\Leftrightarrow x^2+2.x.\dfrac{3}{2}+\dfrac{9}{4}-\dfrac{21}{4}=0\)
\(\Leftrightarrow\left(x+\dfrac{3}{2}\right)^2=\dfrac{21}{4}\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{-3+\sqrt{21}}{2}\\x=\dfrac{-3-\sqrt{21}}{2}\end{matrix}\right.\)
Vậy: \(\Leftrightarrow\left[{}\begin{matrix}x=1\\x=-3\\x=\dfrac{-3+\sqrt{21}}{2}\\x=\dfrac{-3-\sqrt{21}}{2}\end{matrix}\right.\)
