Ta có: 3(x+1)=x+6
\(\Leftrightarrow3x+3-x-6=0\)
\(\Leftrightarrow2x-3=0\)
\(\Leftrightarrow2x=3\)
hay \(x=\dfrac{3}{2}\)
Vậy: \(x=\dfrac{3}{2}\)
3(\(x\)+1)=x+6
⇒3\(x\)+3=\(x\)+6
⇒3\(x\)+3-\(x\)+6=0
⇒2\(x\)-3=0
⇒2\(x\)=3
\(x\)=\(\dfrac{3}{2}\)
vậy \(x=\dfrac{3}{2}\)