5. Find all values of the parameter $a$ for which the equation

$$
\left(\left|\frac{a x^{2}-a x-12 a+x^{2}+x+12}{a x+3 a-x-3}\right|-a\right) \cdot|4 a-3 x-19|=0
$$

has one solution. In your answer, write the largest value of the parameter $a$
Solution:

Simplify $\frac{a x^{2}-a x-12 a+x^{2}+x+12}{a x+3 a-x-3}=\frac{a x^{2}-x^{2}+(-a x+x)+(-12 a+12)}{a(x+3)-(x+3)}=$

$=\frac{x^{2}(a-1)-x(a-1)-12(a-1)}{(x+3)(a-1)}=\frac{(a-1)\left(x^{2}-x-12\right)}{(x+3)(a-1)}=$

$=\frac{x^{2}+3 x-4 x-12}{x+3}=\frac{(x+3)(x-4)}{x+3}=x-4$

We get $\left\{\begin{array}{c}(|x-4|-a) \cdot|4 a+3 x-19|=0 \\ a \neq 1 \\ x \neq-3\end{array} ;\left\{\begin{array}{c}{\left[\begin{array}{c}|x-4|-a=0 \\ 4 a+3 x-19=0 \\ a \neq 1 \\ x \neq-3\end{array} ;\right.} \\ \text {; }\end{array}\right.\right.$

$$
\left\{\begin{array}{c}
a=|x-4| \\
a=-0.75 x+4.75 \\
\quad a \neq 1 \\
x \neq-3
\end{array}\right.
$$

Construct the graph in the coordinate system $\chi O a$ $a=-0.75 x+4.75-$ line

| $x$ | -3 | 5 |
| :--- | :--- | :--- |
| $y$ | 7 | 1 |

$a=|x-4|$ the graph is obtained from $a=|x|$ by shifting 4 units to the right

Since $x \neq-3$, then $a \neq 7$

$a \neq 1$, then $x \neq 3 ; x \neq 5$

Points $(-3 ; 7) ;(3 ; 1) ;(5 ; 1)$ - are excluded

For $a \in(-\infty ; 0) \cup\{7\}$ - one solution

For $a \in(0 ; 1) \cup(1 ; 7) \cup(7 ;+\infty)$ - three solutions

For $a=0$ two solutions

For $a=1$ no solutions

Answer: 7

![](https://cdn.mathpix.com/cropped/2024_05_06_166d4fdfa0c802c9882dg-2.jpg?height=693&width=785&top_left_y=316&top_left_x=1132)

## №3: Triangles.