Ta có : \(\left(x-3\right).\left(x-5\right)=0\)
\(\Leftrightarrow\left[\begin{array}{nghiempt}x-3=0\\x-5=0\end{array}\right.\)
\(\Leftrightarrow\left[\begin{array}{nghiempt}x=0+3\\x=0+5\end{array}\right.\)
\(\Leftrightarrow\left[\begin{array}{nghiempt}x=3\\x=5\end{array}\right.\)
Vậy \(x\in\left\{3;5\right\}\)
\(\left(x-3\right).\left(x-5\right)=0\)
\(\Rightarrow\left[\begin{array}{nghiempt}x-3=0\\x-5=0\end{array}\right.\Leftrightarrow\left[\begin{array}{nghiempt}x=3\\x=5\end{array}\right.\)
vậy \(x=3\) hoặc \(x=5\)
\(\left(x-3\right)\left(x-5\right)=0\)
\(\Rightarrow x-3=0,x-5=0\)
\(\Rightarrow x=3,x-5\)
(x-3).(x-5)=0
ta co : x-3=0 →x=0+3→ x=3
x-5=0→x=0+5→x=5
vay x=3 hoac x=5