(x-2).(x+15) = 0
=> x - 2 = 0 hoặc x + 15 = 0
=> x = 0 + 2 hoặc x = 0 - 15
=> x = 2 hoặc x = -15
\(\left(x-2\right).\left(x+15\right)=0\)
\(\Leftrightarrow\left[\begin{array}{nghiempt}x-2=0\\x+15=0\end{array}\right.\)
\(\Leftrightarrow\left[\begin{array}{nghiempt}x=2\\x=-15\end{array}\right.\)
Vậy \(x\in\left\{2;-15\right\}\)
(x-2). (x +15)=0
x-2=0 hoac x+15=0
x=0+2hoac x =0-15
x = 2 hoac x = -15
vay x=2 hoac -15
x.2 =0 hoặc x+15=0
x =0:2 x =0-15
x =0 x = -15
Vậy x = 0 ; x = -15
