b. \(x^3+4x=0\)
\(\Leftrightarrow x\left(x^2+4\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x^2+4=0\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}x=0\\x^2=-4\end{matrix}\right.\) \(\Leftrightarrow x=0\) ( vì \(x^2\ge0\forall x\) )
Vậy \(x=0\)
\(x^3+4x=0\)
\(\Leftrightarrow x\left(x^2+4\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x=0\\x^2=-4\left(VL\right)\end{matrix}\right.\)
\(\Leftrightarrow x=0\)
Vậy x=0
Ta có: \(x^3+4x=0\)
\(\Leftrightarrow x\left(x^2+4\right)=0\)
hay x=0
x3+4x=0
<=>x(x2+4)=0
<=>\(\left[{}\begin{matrix}x=0\\x^2+4=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\left(tm\right)\\x^2=-4\left(ktm\right)\end{matrix}\right.\)
Vậy x=0
