
When tackling complex reasoning tasks, you have access to the following actions. Use them as needed to progress through your thought process.

[ASSESS]

[ADVANCE]

[VERIFY]

[SIMPLIFY]

[SYNTHESIZE]

[PIVOT]

[OUTPUT]

You should strictly follow the format below:

[ACTION NAME]

# Your action step 1

# Your action step 2

# Your action step 3

...

Next action: [NEXT ACTION NAME]


system
Transform the expression:

\[ \frac{x^{3}+x^{2}-4 x-4}{(x-1)(x-3)+3 x-5}=\frac{(x+1)\left(x^{2}-4\right)}{x^{2}-x-2}=\frac{(x+1)(x-2)(x+2)}{(x-2)(x+1)}\]

\[ 
\begin{gathered}
f(x)=\left|\frac{x^{3}+x^{2}-4 x-4}{(x-1)(x-3)+3 x-5}\right|=|x+2| \\
D(f): x \neq 2, x \neq -1
\end{gathered}
\]

\[ 
p(x)=\sqrt{x^{2}}+a=|x|+a \text{, obtain the equation: } |x+2|=|x|+a
\]

Solve it graphically in the coordinate system XOA:

\[ 
\begin{gathered}
\left\{\begin{array}{c}
x<-2 \\
-x-2=-x+a \\
\end{array}\right. \Leftrightarrow \left\{\begin{array}{l}
x<-2 \\
a=-2 \\
\end{array}\right. 
\\
\left\{\begin{array}{c}
-2 \leq x<0 \\
x+2=-x+a \\
\end{array}\right. 
\Leftrightarrow \left\{\begin{array}{c}
-2 \leq x<0 \\
a=2x+2 \\
\end{array}\right. 
\\
\left\{\begin{array}{c}
x \geq 0 \\
x+2=x+a \\
\end{array}\right. 
\Leftrightarrow \left\{\begin{array}{c}
x \geq 0 \\
a=2x+2 \\
\end{array}\right. 
\end{gathered}
\]

The equation has 1 solution when:

\[ a \in (-2, 0) \cup (0, 2) \]

Present the answer in LaTex format: \boxed{Your answer}
user