$1^{0}$. Which is the quadratic equation whose roots, $x_{1}$ and $x_{2}$, are real and satisfy the following relationships:

$$
\begin{array}{r}
x_{1}+x_{2}+x_{1} x_{2}=2 m+k \ldots \\
\left(x_{1}-1\right)\left(x_{2}-1\right)=m+1-k \ldots
\end{array}
$$

$2^{0}$. If $x_{1}-x_{2}=1$, then there is a relationship between $k$ and $m$ such that the roots of the aforementioned equation are real.

$3^{0}$. What are $k$ and $m$ if, in addition to the above, $k-m=1$? Calculate $x_{1}$ and $x_{2}$.
A $1^{0}$. The sought equation should be: $x^{2}+p x+q=0$, where therefore

$$
x_{1}+x_{2}=-p \quad \text { and } \quad x_{1} x_{2}=q \text {. }
$$

Considering this, from 1) and 2)

$$
\begin{aligned}
-p+2 q & =2 m+k \ldots \\
p+q+1 & =m+1-k \ldots
\end{aligned}
$$

From here:

$$
3 q=3 m, \quad \text { that is } \quad q=m \quad \text { and thus } \quad p=-k \text {. }
$$

The sought equation is:

$$
x^{2}-k x+m=0 \ldots
$$

The roots of this equation are real if $k^{2} \geq 4 m$.

$2^{0}$. If $x_{1}-x_{2}=1$, then, since $x_{1}+x_{2}=k$,

$$
x_{1}=\frac{1+k}{2} \quad \text { and } x_{2}=\frac{k-1}{2}, \quad x_{1} x_{2}=\frac{k^{2}-1}{4} \text {. }
$$

However, $x_{1} x_{2}=q=m$, so $k^{2}-1=4 m, k^{2}>4 m$.

$3^{0}$. According to the above, $k^{2}=4 m+1$. If additionally $k-m=1$, that is $k=m+1$, then

$$
(m+1)^{2}=4 m+1 \text { and thus } m^{2}-2 m=m(m-2)=0 \text {. }
$$

Therefore

$$
m_{1}=0 \quad \text { and } k_{1}=1, \quad \text { or } \quad m_{2}=2 \text { and } k_{2}=3 .
$$

For $k_{1}=1$

$$
\begin{aligned}
& x_{1}=1 \quad \text { and } \quad x_{2}=0 . \\
& x_{1}=2 \quad \text { and } \quad x_{2}=1 .
\end{aligned}
$$

For $k_{2}=3$