\(\left|x-2\right|=3\)
TH1: \(x-2=3\)
\(x=3+2\)
\(x=5\)
TH2: \(x-2=-3\)
\(x=\left(-3\right)+2\)
\(x=-1\)
Vậy \(x=\left\{-1;5\right\}\)
|x-2|=3
=> x-2 = 3 hoặc x-2 = -3
+ x-2 = 3 <=> x = 5
+ x-2 = -3 <=> x = -1
Vậy x= -1 hoặc x =5
| x-2 | =3
\(\Rightarrow\left[{}\begin{matrix}x-2=3\\x-2=-3\end{matrix}\right.\)\(\Rightarrow\left[{}\begin{matrix}x=3+2\\x=-3+2\end{matrix}\right.\)\(\Rightarrow\left[{}\begin{matrix}x=5\\x=-1\end{matrix}\right.\)
Vậy x=5 hoặc x=-1
\(|x-2|=3\)
\(\Rightarrow x-2=3\) hoặc \(x-2=-3\)
\(\Rightarrow x=5\) hoặc \(x=-1\)
Vậy \(x\in\left\{5;-1\right\}\)