4. Find all values of the parameter $a$, for each of which the equation $a|x+1|+\frac{x^{2}-5 x+6}{2-x}=0$ has exactly one solution
Answer: $a \in\left[-1 ;-\frac{1}{3}\right) \cup\left(-\frac{1}{3} ; 1\right)$.

Solution. Given the condition $x \neq 2$, the equation is equivalent to $a|x+1|=x-3$. The graph of the right-hand side of the equation is the line $y=x-3$. The graph of the left-hand side of the equation is a "V" shape with its vertex at the point $(-1 ; 0)$, and the slopes of its branches are determined by the parameter $a$.

The left branch of the "V" intersects the line when $a<-1$, and the right branch intersects the line when $a<1$ and $a \neq-\frac{1}{3}$ (when $a=-\frac{1}{3}$, the right branch passes through the punctured point of the line - the point $(2 ;-1)$). Therefore, exactly one solution is obtained when $a \in\left[-1 ;-\frac{1}{3}\right) \cup\left(-\frac{1}{3} ; 1\right)$.