9. Specify all values of $a$ for which the system of equations $y=\frac{x+|x|}{x} ;(x-a)^{2}=y+a$ has at least one solution, and solve it for each $a$.

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# Solution:

I. When $x>0 \quad y=2, \quad x^{2}-2 a x+a^{2}-a-2=0(*)$. Equation $(*)$ has two distinct positive roots $x_{1,2}=a \pm \sqrt{a+2}$, if: $\left\{\begin{array}{c}D / 4=a+2>0, \\ a>0, \\ a^{2}-a-2>0\end{array} \Leftrightarrow\left\{\begin{array}{c}a>-2, \\ a>0, \\ {\left[\begin{array}{c}a2\end{array}\right.}\end{array} \Leftrightarrow a>2\right.\right.$. Equation (*) has one positive root $x_{1,2}=a+\sqrt{a+2}$, if: $\left[\begin{array}{c}\left\{\begin{array}{l}D=0, \\ a>0,\end{array}\right. \\ a^{2}-a-20, \\ a0 .\end{array}\right.$ has no solutions. Equation (**) has one negative root $x=a-\sqrt{a}$, if

$$
\left[\begin{array}{l}
\left\{\begin{array}{l}
D / 4=a=0, \\
\quad a<0,
\end{array}\right. \\
\left\{\begin{array}{l}
a^{2}-a<0, \\
\left\{\begin{array}{l}
a=0, \\
a=1 \\
a<0
\end{array}\right.
\end{array} \Leftrightarrow 0<a<1\right.
\end{array}\right.
$$

Answer: when $a \in(-1 ; 0] \cup[1 ; 2] \quad x=a+\sqrt{a+2}, y=2$;

when $(0 ; 1) x_{1}=a+\sqrt{a+2}, y_{1}=2 ; x_{2}=a-\sqrt{a} ; y_{2}=0$;

when $a \in(2 ;+\infty) x_{1,2}=a \pm \sqrt{a+2}, y_{1,2}=2$.