## Task B-2.2.

Determine all real numbers $m$ such that for the solutions $x_{1}, x_{2}$ of the equation $x^{2}+m-3 x=m x-2$, the inequality $\frac{x_{1}}{x_{1}+1}+\frac{x_{2}}{x_{2}+1}<1$ holds.
## Solution.

We write the equation $x^{2}+m-3 x=m x-2$ in the form $x^{2}-(m+3) x+m+2=0$. The left side of the given inequality can be written as

$$
\frac{x_{1}}{x_{1}+1}+\frac{x_{2}}{x_{2}+1}=\frac{x_{1}\left(x_{2}+1\right)+x_{2}\left(x_{1}+1\right)}{\left(x_{1}+1\right)\left(x_{2}+1\right)}=\frac{2 x_{1} x_{2}+x_{1}+x_{2}}{x_{1} x_{2}+x_{1}+x_{2}+1}
$$

According to Vieta's formulas, we have:

$$
x_{1}+x_{2}=m+3, \quad x_{1} x_{2}=m+2
$$

1 point

From $\frac{2 x_{1} x_{2}+x_{1}+x_{2}}{x_{1} x_{2}+x_{1}+x_{2}+1}<1$ it follows that

$$
\begin{gathered}
\frac{2(m+2)+(m+3)}{(m+2)+(m+3)+1}<1 \\
\frac{3 m+7}{2 m+6}<1 \\
\frac{m+1}{2 m+6}<0
\end{gathered}
$$

This inequality is equivalent to the inequality $(m+1)(2 m+6)<0$ whose solution is $m \in(-3,-1)$.