Phương trình $\dfrac{{x - 1}}{2} + \dfrac{{x - 1}}{3} - \dfrac{{x - 1}}{6} = 2$ có tập nghiệm là

$\dfrac{x}{7} + 3 = 0$

$(x - 1)(x + 2) = 0$

$15 - 6x = 3x + 5$

$x = 3x + 2$

${x_0} = 4$                       

${x_0} < 4$                     

${x_0} > 4$                 

${x_0} > 5$       

$7,2$

$3,6$

$14,4$

$1,8$

$\dfrac{{AD}}{{AB}} = \dfrac{{AE}}{{AC}} \Rightarrow DE//BC$.

$\dfrac{{AD}}{{DB}} = \dfrac{{AE}}{{EC}} \Rightarrow DE//BC$.

$\dfrac{{AB}}{{DB}} = \dfrac{{AC}}{{EC}} \Rightarrow DE//BC$.

$\dfrac{{AD}}{{DE}} = \dfrac{{AE}}{{ED}} \Rightarrow DE//BC$.

$\dfrac{1}{2}$                       

$- \dfrac{1}{2}$                      

$0$

$2$

$x = 9$

$x =  - 9$

$x = 8$

$x =  - 8$

$7$         

$\dfrac{1}{2}$          

$\dfrac{7}{4}$          

$\dfrac{7}{{16}}$