a,x^2+3x=0
=> x.(x+3)=0
=> +)x=0
+) x+3=0 => x=-3
b,x^3-4x=0
=> x.(x^2-2^2)=0
=> x.(x-2).(x+2)=0
=> +) x=0
+) x-2=0 => x=2
+) x+2=0 => x= -2
a) \(x^2+3x=0\)
\(x\left(x+3\right)=0\)
\(\Rightarrow\orbr{\begin{cases}x=0\\x+3=0\end{cases}}\Rightarrow\orbr{\begin{cases}x=0\\x=-3\end{cases}}\)
vay \(\orbr{\begin{cases}x=0\\x=-3\end{cases}}\)
b) \(x^3-4x=0\)
\(x\left(x^2-4\right)=0\)
\(\Rightarrow\orbr{\begin{cases}x=0\\x^2-4=0\end{cases}}\Rightarrow\orbr{\begin{cases}x=0\\x^2=4\end{cases}}\Rightarrow\orbr{\begin{cases}x=0\\x=2\end{cases}}\)
vay \(\orbr{\begin{cases}x=0\\x=2\end{cases}}\)
a) \(x^2+3x=0\)
\(x\left(x+3\right)=0\)
\(\Rightarrow\orbr{\begin{cases}x=0\\x+3=0\end{cases}}\)
\(\Rightarrow\orbr{\begin{cases}x=0\\x=-3\end{cases}}\)
Vậy \(x\in\left\{-3;0\right\}\)
b) \(x^3-4x\)
\(x\left(x^2-2^2\right)=0\)
\(x\left(x-2\right)\left(x+2\right)=0\)
=> x = 0
Hoặc x - 2 = 0 => x = 2
Hoặc x + 2 = 0 => x = -2
Vậy \(x\in\left\{-2;0;2\right\}\)
a) \(x^2+3x=0\)
\(x\left(x+3\right)=0\)
\(\Rightarrow\orbr{\begin{cases}x=0\\x+3=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=0\\x=0-3\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=0\\x=-3\end{cases}}\)
b) \(x^3-4x=0\)
\(x\left(x^2-4\right)=0\)
\(\Rightarrow\orbr{\begin{cases}x=0\\x^2-4=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=0\\x^2=4\end{cases}\Leftrightarrow\orbr{\begin{cases}x=0\\x=\pm2\end{cases}}}\)
