\(\left(x-3\right)\left(x-1\right)\left(x+1\right)\left(x+3\right)+15=0\)
\(\Leftrightarrow\left(x^2-1\right)\left(x^2-9\right)+15=0\)
Đặt \(x^2-1=t\)nên \(pt\Leftrightarrow t\left(t-8\right)+15=0\)
\(\Leftrightarrow t^2-8t+15=0\)
\(\Leftrightarrow t^2-3t-5t+15=0\)
\(\Leftrightarrow t\left(t-3\right)-5\left(t-3\right)=0\)
\(\left(t-5\right)\left(t-3\right)=0\Rightarrow\orbr{\begin{cases}t-5=0\\t-3=0\end{cases}\Rightarrow\orbr{\begin{cases}t=5\\t=3\end{cases}}}\)
Với \(t=5\) thì \(x^2-1=5\Leftrightarrow x^2=6\Rightarrow\orbr{\begin{cases}x=\sqrt{6}\\x=-\sqrt{6}\end{cases}}\)
Với \(t=3\) thì \(x^2-1=3\Leftrightarrow x^2=4\Rightarrow\orbr{\begin{cases}x=-2\\x=2\end{cases}}\)
Vậy \(x\in\left\{-\sqrt{6};-2;2\sqrt{6}\right\}\)
(x-3)x-1)(x+1)(x+3)+15=0
<=>(x^2-9)(x^2-1)=-15
<=>x^4-10x^2-9=-15
<=>-(x^4+6x^2+9+4X^2)=-15
