\(3x+\left(x-2\right)=2x+1\)
\(\Leftrightarrow3x+x-2=2x+1\)
\(\Leftrightarrow3x+x-2x=1+2\)
\(\Leftrightarrow2x=3\)
\(\Leftrightarrow x=\dfrac{3}{2}\)
Vậy \(x=\dfrac{3}{2}\)
\(3x+\left(x-2\right)=2x+1\)
\(\Leftrightarrow3x+x-2=2x+1\)
\(\Leftrightarrow3x+x-2x=2+1\)
\(\Leftrightarrow2x=4\)
\(\Leftrightarrow x=\dfrac{4}{2}\)
\(\Leftrightarrow x=2\)
Vậy \(x=2\)
\(3x+\left(x-2\right)=2x+1\)
\(\Rightarrow3x+x-2=2x+1\)
\(\Rightarrow4x-2=2x+1\)
\(\Rightarrow4x-2x=2+1\)
\(\Rightarrow2x=3\)
\(\Rightarrow x=\dfrac{3}{2}\)
