Đặt \(\left\{{}\begin{matrix}x^2-4=a\\x^2-5x+1=b\end{matrix}\right.\) \(\Rightarrow\frac{a-b}{5}=x-1\) pt trở thành:
\(a.b=6\left(\frac{a-b}{5}\right)^2\Leftrightarrow25ab=6a^2-12ab+6b^2\)
\(\Leftrightarrow6a^2-37ab+6b^2=0\Leftrightarrow6a^2-ab-36ab+6b^2=0\)
\(\Leftrightarrow a\left(6a-b\right)-6b\left(6a-b\right)=0\Leftrightarrow\left(a-6b\right)\left(6a-b\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}a=6b\\6a=b\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}x^2-4=6\left(x^2-5x+1\right)\\6\left(x^2-4\right)=x^2-5x+1\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}5x^2-30x+10=0\\5x^2+5x-25=0\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}x=3\pm\sqrt{7}\\x=\frac{-1\pm\sqrt{21}}{2}\end{matrix}\right.\)
