Ta có: \(\frac{x-1}{2}=\frac{2x-1}{3}\Rightarrow3\left(x-1\right)=2\left(2x-1\right)\)
\(\Leftrightarrow3x-3=4x-2\)
\(\Leftrightarrow3x-4x=-2+3\)
\(\Leftrightarrow-x=1\)
Vậy x = -1
\(x-\frac{1}{2}=2x-\frac{1}{3}\)
\(\frac{1}{2}=x-\frac{1}{3}\)Loại x ở hai vế
\(x=\frac{1}{3}+\frac{1}{2}=\frac{5}{6}\)
\(x-\frac{1}{2}=2x-\frac{1}{3}\)
\(\frac{1}{2}=x-\frac{1}{3}\)
\(x=\frac{1}{2}+\frac{1}{3}\)
\(=\frac{5}{6}\)
\(x-\frac{1}{2}=2x-\frac{1}{3}\)
\(\left(x-\frac{1}{2}\right)-\left(2x-\frac{1}{3}\right)=0\)
\(x-\frac{1}{6}-2x=0\)
\(x-\frac{1}{6}=2x\)
\(-\frac{1}{6}=x\)
Nhầm dấu mất,xin phép được giải lại
