Giải phương trình sau 2(x + 1) / 3 = 1/2 - (1 - 2x) / 6

\(\frac{{2(x + 1)}}{3} = \frac{1}{2} - \frac{{1 - 2{\rm{x}}}}{6}\)       

     \(\frac{{4(x + 1)}}{6} = \frac{3}{6} - \frac{{1 - 2{\rm{x}}}}{6}\)                                

         4x + 4 = 3 – 1 + 2x                      

\(\)       4x – 2 = 2 − 4                                                  

                2x = −2                                                                    

                  x =−1      

Vậy phương trình có nghiệm x = −1