`|x-2|=3-x`
`@TH1:x-2 >= 0<=>x >= 2=>|x-2|=x-2`
`=>x-2=3-x`
`<=>2x=5`
`<=>x=5/2` (t/m)
`@TH2:x-2 < 0<=>x < 2=>|x-2|=2-x`
`=>2-x=3-x`
`<=>0x=1` (Vô lí)
Vậy `S={5/2}`
\(\left|x-2\right|=3-x\)
\(\Leftrightarrow\left[{}\begin{matrix}x-2=3-x\\x-2=x-3\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x-2-3+x=0\\x-2-x+3=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}2x-5=0\\\left(x-x\right)+\left(-2+3\right)=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{5}{2}\\1=0\left(vl\right)\end{matrix}\right.\)
\(=>x=\dfrac{5}{2}\)
+TH1:x−2≥0⇔x≥2⇒|x−2|=x−2∘TH1:x-2≥0⇔x≥2⇒|x-2|=x-2
⇒x−2=3−x⇒x-2=3-x
⇔2x=5⇔2x=5
⇔x=52⇔x=52 (t/m)
+TH2:x−2<0⇔x<2⇒|x−2|=2−x∘TH2:x-2<0⇔x<2⇒|x-2|=2-x
⇒2−x=3−x⇒2-x=3-x
⇔0x=1⇔0x=1 (Vô lí)
vậy S={5/2}
