\(\left(3-x\right)\left(x^2+1\right)=0\)
\(\Rightarrow x=3\)(do \(x^2+1\ge1>0\forall x\))
\(\left(3-x\right)\left(x^2+1\right)=0\\ \Rightarrow\left[{}\begin{matrix}3-x=0\\x^2+1=0\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}x=3\\x^2=-1\left(vôlí\right)\end{matrix}\right.\)
Vậy \(x=3\)
Vì `x^2 >=0\ forall x -> x^2+1 >= 1 >0 `
` -> 3-x=0`
`-> x=3`.
Ta có: \(x^2+1>0dox^2\ge0\) với mọi x.
Khi đó: \(3-x=0\Rightarrow x=3\)
