`a)x(x-3)-x^2=2`
`<=>x^2-3x-x^2=2`
`<=>-3x=2`
`<=>x=-2/3`
_______________________________
`b)x^3-16x=0`
`<=>x(x^2-16)=0`
`<=>x(x-4)(x+4)=0`
`<=>[(x=0),(x=4),(x=-4):}`
\(a,x\left(x-3\right)-x^2=2\\ \Rightarrow x^2-3x-x^2=2\\ \Rightarrow-3x=2\\ \Rightarrow x=-\dfrac{2}{3}\\ -----\\ b,x^3-16x=0\\ \Rightarrow x\left(x^2-16\right)=0\\ \Rightarrow\left[{}\begin{matrix}x=0\\x-4=0\\x+4=0\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}x=0\\x=4\\x=-4\end{matrix}\right.\)
`x.(x-3)-x^2 = 2`
`x^2 - 3x - x^2 = 2`
`-3x=2`
`x=-2/3`
_____________________________
`x^3 - 16x=0`
`x.(x+4).(x-4)=0`
`=>@Th1:`
`x=0`
`@Th2:`
`x+4=0`
`x=0-4`
`x=-4`
`@Th3:
`x-4=0`
`x=0+4`
`x=4`
a, \(x\left(x-3\right)-x^2=2\)
\(\Leftrightarrow x^2-3x-x^2=2\)
\(\Leftrightarrow3x=2\)
\(\Leftrightarrow x=\dfrac{2}{3}\)
b, \(x^3-16x=0\)
\(\Leftrightarrow x\left(x^2-16\right)=0\)
\(\Leftrightarrow x\left(x-4\right)\left(x+4\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x-4=0\\x+4=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=4\\x=-4\end{matrix}\right.\)
\(a,
x\left(x-3\right)-x^2=2\)
\(x^2-3x-x^2=2\)
\(-3x=2\)
Vậy \(x=-\dfrac{2}{3}\)
_________________________
\(b,
x^3-16x=0\)
\(x\left(x^2-16\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x=0\\x^2-16=0\end{matrix}\right.
\Rightarrow\left[{}\begin{matrix}x=0\\x=4\end{matrix}\right.\)
Vậy \(x\in\left\{0;
4\right\}\)
