a) \(\left(3-x\right)\left(x-3\right)=0\)
\(\Leftrightarrow x-3=0\)
\(\Leftrightarrow x=3\)
b) \(x\left(x+1\right)=0\)
\(\Rightarrow x\in\left\{0;-1\right\}\)
a, ( 3 - x ).( x - 3 ) = 0
\(\orbr{\begin{cases}3-x=0\\x-3=0\end{cases}}\)\(\Leftrightarrow\)\(\orbr{\begin{cases}x=3\\x=3\end{cases}}\)
Vậy x = 3
b, x.( x + 1 ) = 0
\(\orbr{\begin{cases}x=0\\x+1=0\end{cases}}\)\(\Leftrightarrow\)\(\orbr{\begin{cases}x=0\\x=-1\end{cases}}\)
Vậy x = 0 hoặc x = - 1
