Ta có : \(\left(x-1\right)\left(x-2\right)\left(x^2+20\right)=0\)
\(\Leftrightarrow\left[\begin{array}{nghiempt}x-1=0\\x-2=0\\x^2+20=0\end{array}\right.\)
\(\Leftrightarrow\left[\begin{array}{nghiempt}x=1\\x=2\\x^2=-20\left(loại\right)\end{array}\right.\)
Vậy \(x\in\left\{1;2\right\}\)
Ta có: (x-1)(x-2)(x2+20)=0
\(\Rightarrow\begin{cases}x-1=0\\x-2=0\\x^2+20=0\end{cases}\Rightarrow\begin{cases}x=1\\x=2\\\varnothing\end{cases}}\)
vậy x=1 hoặc x=2