9. For what values of the parameter a does the equation $f(x)=p(x)$ have one solution, if $f(x)=$ $\left|\frac{2 x^{3}-5 x^{2}-2 x+5}{(1.5 x-3)^{2}-(0.5 x-2)^{2}}\right|, \quad p(x)=|2 x+5|+a$. If there are more than one value of the parameter, then in the answer, specify their sum.

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# Solution:

Simplify $f(x)=\left|\frac{2 x^{3}-5 x^{2}-2 x+5}{(1.5 x-3)^{2}-(0.5 x-2)^{2}}\right|$, we get

$$
f(x)=|x+1|, \text { where } x \neq 1, x \neq 2.5 \text {. }
$$

Solve the equation $|x+1|=|5+2 x|+a$, where $x \neq 1, x \neq 2.5$ graphically in the system $x O a$.

1) $\left\{\begin{array}{c}x<-2.5 \\ -x-1=-2 x-5+a\end{array} \quad\left\{\begin{array}{l}x<-2.5 \\ a=x+4\end{array}\right.\right.$

![](https://cdn.mathpix.com/cropped/2024_05_06_a3feddd7c418bd26b633g-8.jpg?height=634&width=649&top_left_y=748&top_left_x=1246)

2) $\left\{\begin{array}{c}-2.5 \leq x<-1 \\ -x-1=2 x+5+a\end{array} \quad\left\{\begin{array}{c}-2.5 \leq x<-1 \\ a=-3 x-6\end{array}\right.\right.$
3) $\left\{\begin{array}{c}x \geq-1 \\ x+1=2 x+5+a\end{array} \quad\left\{\begin{array}{c}x \geq-1 \\ a=-x-4\end{array}\right.\right.$

The equation has one solution when $a=-6.5, a=-5, a=1.5$. The sum is -10.

Answer: -10.