\(\Leftrightarrow x\left(x+5\right)\left(x+2\right)\left(x+3\right)=280\)
\(\Leftrightarrow\left(x^2+5x\right)\left(x^2+5x+6\right)=280\)
Đặt \(x^2+5x=t\)
\(\Rightarrow t\left(t+6\right)=280\)
\(\Leftrightarrow t^2+6t-280=0\)
\(\Leftrightarrow t^2+20t-14t-280=0\)
\(\Leftrightarrow t\left(t+20\right)-14\left(t+20\right)=0\)
\(\Leftrightarrow\left(t-14\right)\left(t+20\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}t=14\\t=-20\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x^2+5x=14\\x^2+5x=-20\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x^2+5x-14=0\\x^2+5x+20=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}\left(x+7\right)\left(x-2\right)=0\\\left(x+\dfrac{5}{2}\right)^2+\dfrac{55}{4}=0\left(vn\right)\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=-7\\x=2\end{matrix}\right.\)
