\(\left(x+a\right)\left(x+4a\right)\left(x+2a\right)\left(x+3a\right)=3a^4\)
\(\Leftrightarrow\left(x^2+5ax+4a^2\right)\left(x^2+5ax+6a^2\right)-3a^4=0\)
Đặt \(x^2+5ax+4a^2=t\)
\(\Rightarrow t\left(t+2a^2\right)-3a^4=0\Leftrightarrow t^2+2a^2t-3a^4=0\)
\(\Leftrightarrow\left(t-a^2\right)\left(t+3a^2\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}t=a^2\\t=-3a^2\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}x^2+5ax+4a^2=a^2\\x^2+5ax+4a^2=-3a^2\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x^2+5ax+3a^2=0\left(1\right)\\x^2+5ax+7a^2=0\left(2\right)\end{matrix}\right.\)
(1) \(\Rightarrow x=\frac{-5a\pm a\sqrt{13}}{2}\)
(2): nếu \(a=0\Rightarrow x=0\), nếu \(a\ne0\Rightarrow\) (2) vô nghiệm
