\(x^3\left(x-5\right)+3x\left(x-5\right)=0\)
\(x^4-5x^3+3x^2-15x=0\)
Chịu!!!!....
\(x^3\left(x-5\right)+3x\left(x-5\right)=0\Leftrightarrow\left(x^3+3x\right)\left(x-5\right)=0\Leftrightarrow x\left(x^2+3\right)\left(x-5\right)=0\)
\(\Leftrightarrow\hept{\begin{cases}x=0\\x^2+3=0\\x-5=0\end{cases}}\) ( vô lí vì \(x^2+3\ge3>0\) với mọi x)
Suy ra \(\hept{\begin{cases}x=0\\x=5\end{cases}}\)