Решение:
$( a + 2 x ) \sqrt{1 + 2 ax} = ( a + 2 x ) ( a + x + 1 )$
$( a + 2 x ) ( \sqrt{1 + 2 ax} - a - x - 1 ) = 0$
$\left[\begin{matrix} \left\{\begin{matrix} \begin{matrix}a + 2 x = 0 , \\ 1 + 2 ax \geq 0 \Leftrightarrow a \geq - \frac{1}{2 x}\end{matrix} \end{matrix}\right. \\ a + x + 1 = \sqrt{1 + 2 ax} , ( a \geq - x - 1 ) \end{matrix}\right.$
$\left(( a + x )\right)^{2} + 2 ( a + x ) + 1 = 1 + 2 ax$
$a^{2} + 2 ax + x^{2} + 2 a + 2 x + 1 = 1 + 2 ax$
$a^{2} + 2 a + x^{2} + 2 x = 0$
$\left\{\begin{matrix} \left(( a + 1 )\right)^{2} + \left(( x + 1 )\right)^{2} = 2 , \\ a \geq - x - 1 \end{matrix}\right.$

1) $\left\{\begin{matrix} x = - a - 1 , \\ \left(( a + 1 )\right)^{2} + \left(( x + 1 )\right)^{2} = 2 \end{matrix}\right.$
$2 a^{2} + 2 a - 1 = 0 \\ a = - \frac{1}{2} \pm \frac{\sqrt{3}}{2} .$
2) $\left\{\begin{matrix} a = - 2 x \Rightarrow x = - \frac{1}{2} a , \\ \left(( a + 1 )\right)^{2} + \left(( x + 1 )\right)^{2} = 2 \end{matrix}\right.$
$a_{1} = 0 , a_{2} = - \frac{4}{5} .$
3) $a = 0 .$
$4 ) a \in [ - \frac{1}{2} + \frac{\sqrt{3}}{2} , - 1 + \sqrt{2} ) .$

Ответ: $a \in [ - \frac{1}{2} - \frac{\sqrt{3}}{2} , - \frac{4}{5} ) \cup ( - \frac{4}{5} ; 0 ) \cup ( 0 , - \frac{1}{2} + \frac{\sqrt{3}}{2} ) \cup \left\{\right. - 1 + \sqrt{2} \left.\right\}$

Источник: ФИПИ