\(\left(x^2+2x-3\right)\left(x^2+2x-4\right)=2\) (1)
Đặt: \(x^2+2x-3=t\)
PT (1) <=> t. (t - 1) = 2
\(\Leftrightarrow t^2-t-2=0\)
\(\Leftrightarrow t^2+t-2t-2=0\)
\(\Leftrightarrow t\left(t+1\right)-2\left(t+1\right)=0\)
\(\Leftrightarrow\left(t+1\right)\left(t-2\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}t+1=0\\t-2=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}t=-1\\t=2\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x^2+2x-3=-1\\x^2+2x-3=2\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x^2+2x-2=0\\x^2+2x-5=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-1-\sqrt{3}\\x=-1+\sqrt{3}\\x=-1+\sqrt{6}\\x=-1-\sqrt{6}\end{matrix}\right.\)
