\(\left(x^2-9\right)^2=12x+1\)
\(\Leftrightarrow x^4-18x^2+81-12x-1=0\)
\(\Leftrightarrow x^4-2x^3+2x^3-4x^2-14x^2+28x-40x+80=0\)
\(\Leftrightarrow x^3\left(x-2\right)+2x^2\left(x-2\right)-14x\left(x-2\right)-40\left(x-2\right)=0\)
\(\Leftrightarrow\left(x-2\right)\left(x^3+2x^2-14x-40\right)=0\)
\(\Leftrightarrow\left(x-2\right)\left(x^3-4x^2+6x^2-24x+10x-40\right)=0\)
\(\Leftrightarrow\left(x-2\right)\left[x^2\left(x-4\right)+6x\left(x-4\right)+10\left(x-4\right)\right]=0\)
\(\Leftrightarrow\left(x-2\right)\left(x-4\right)\left(x^2+6x+10\right)=0\) (*)
Vì\(x^2+6x+10=x^2+6x+9+1=\left(x+3\right)^2+1>0\)
(*) \(\Leftrightarrow\left(x-2\right)\left(x-4\right)=0\Leftrightarrow\left[{}\begin{matrix}x-2=0\\x-4=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=2\\x=4\end{matrix}\right.\)
Vậy tập nghiệm của phương trình là S = {2;4}.
\(\left(x^2-9\right)^2=12x+1\\ \Leftrightarrow\left(x^29\right)^2-1=12x\\ \Leftrightarrow\left(x^2-10\right)\left(x^2-8\right)-12x=0\\ \Leftrightarrow x^4-18x^2-12x+80=0\\ \Leftrightarrow x^4-4x^3+4x^3-16x^2-2x^2+8x-20x+80=0\\ \Leftrightarrow\left(x^3+4x^2-2x-20\right)\left(x-4\right)=0\\ \Leftrightarrow\left(x^3-2x^2+6x^2-12x+10x-20\right)\left(x-4\right)=0\\ \Leftrightarrow\left(x^2+6x+10\right)\left(x-2\right)\left(x-4\right)=0\\ \Leftrightarrow\left(\left(x+3\right)^2+1\right)\left(x-2\right)\left(x-4\right)=0\)
từ đó suy ra x=2 hoac x=4
