\(\left|x-2\right|+3=2x\)
\(\Leftrightarrow\left|x-2\right|=2x-3\)
TH1: x ≥ 2
\(x-2=2x-3\)
\(\Leftrightarrow x=1\left(ktm\right)\)
TH2: x < 2
\(x-2=-2x+3\)
\(\Leftrightarrow x=\dfrac{5}{3}\left(tm\right)\)
Vậy...
Ta có : /x-2/+3=2x ⇒2x-/x-2/=3
nếu x-2 ≥0 ⇒x≥2 ⇒/x-2/=x-2
suy ra 2x-(x-2)=3⇒x=1( ko thoả mãn đk x≥2)
nếu x-2<0⇒x<2⇒/x-2/= - x+2
suy ra 2x-(-x+2)=3⇒x=\(\dfrac{5}{3}\)(tm đk x<2)
Vậy x=\(\dfrac{5}{3}\)