a) \(x^3-16x=0\)
\(\Leftrightarrow x\left(x^2-16\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x=0\\x^2-16=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=0\\x=\pm\sqrt{16}=\pm4\end{cases}}\)
Vậy \(x\in\left\{0;\pm4\right\}\)
b) \(x^2-6x+9=0\)
\(\Leftrightarrow\left(x-3\right)^2=0\)
\(\Leftrightarrow x-3=0\)
\(\Leftrightarrow x=3\)
\(a)x^3-16x=0=>x\left(x^2-16\right)=0\)
\(=>x\left(x-4\right)\left(x+4\right)=0\)
Suy ra x=0 hoặc x-4=0 hoặc x+4=0
=> x=0 hoặc x=4 hoặc x=-4
b)\(x^2-6x+9=0=>\left(x-3\right)^2=0=>x-3=0=>x=3\)
