Question

giải phương trình

3x [x² + 1 ] [ x - 2 ] = 0

Answer 1

\(3x\left(x^2+1\right)\left(x-2\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x^2+1=0\\x-2=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x\in\varnothing\\x=2\end{matrix}\right.\Leftrightarrow x=2\)

Vậy ..........

Answer 2

\(3x\left(x^2+1\right)\left(x-2\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}3x=0\\x^2+1=0\\x-2=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x\in\varnothing\\x=2\end{matrix}\right.\)

Vậy \(x\in\left\{0;2\right\}\)