The solution use Thevenin's theorem to solve the problem,but i can't understand why
where does \$ \frac{−10}{2}\$ come from? I don't understand!


When the author tries to find \$V_X\$ voltage he simply converts the voltage source to a current source.


Also, you should always track the units, in this case, you will see that \$\frac{10V}{2\Omega} = 5A\$ has a unit of amperes.  

$$ V_X = (2A - 5A) \times 4\Omega||2\Omega =-4V $$


  where does 1/2 come from? I don't understand!


Again you don't track the units and this is why you don't see what is going on. 

So, we have this situation: 



And we want to know the voltage difference between two points \$A\$ and \$B\$

We see that the voltage at point \$B\$ is equal to \$-10V\$ with respect to ground.

So, we need to find the voltage at point \$A\$ also with respect to ground. 

But this is not a problem because we already know the voltage at point \$V_X\$. 

The current that will flow via \$R_2 + R_3\$ is equal to: 

$$I = \frac{V_X}{R_2 + R_3} = -1A$$  

And the voltage at point \$A\$ is 

$$V_A = I \times R_3 =\frac{V_X}{R_2 + R_3} \times R_3 = V_X \times \frac{R_3}{R_2 + R_3} = -4V \times \frac{2\Omega}{2\Omega + 2\Omega} = -4V \times \frac{2\Omega}{4\Omega} =$$ 

$$ = -4V \times \frac{1\Omega}{2\Omega} = -2V$$

Therefore: 

$$V_{AB} = V_A - V_B = (-2V) - (-10V) = 8V$$