TH1: \(x-2=5\) \(\Rightarrow x=7\) |
TH2: \(-x+2=5\) \(\Rightarrow x=-3\) |
`|x-2|=5`
`<=>` $\\left[ \begin{array}{l}x-2=5\\x-3=-5\end{array} \right.$
`<=>` $\\left[ \begin{array}{l}x=7\\x=-2\end{array} \right.$
Vậy `s={7,-2}`
|x-2|=5
=>x-2=5 hoặc x-2=-5
+)x-2=5 =>x=7
+)x-2=-5 =>x=-3
Vậy x thuộc {7;-3}
|x-2|=5
TH1: x-2≥0 ⇔ x≥2 TH2: x-2<0 ⇔ x<2
x-2=5 x-2=-5
⇔ x=7(tm) ⇔ x=-3(tm)
⇒S={7;-3}