Hai phương trình tương đương là hai phương trình có

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

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

$15 - 6x = 3x + 5$

$x = 3x + 2$

\({x_0} = 4\)                       

\({x_0} < 4\)                     

\({x_0} > 4\)                 

\({x_0} > 5\)       

\(7,2\)

\(3,6\)

\(14,4\)

\(1,8\)

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

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

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

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

\(\dfrac{1}{2}\)                       

\( - \dfrac{1}{2}\)                      

$0$

$2$

$x = 9$

$x =  - 9$

$x = 8$

$x =  - 8$

$7 $         

\(\dfrac{1}{2}\)          

\(\dfrac{7}{4}\)          

\(\dfrac{7}{{16}}\)