1. Given are the parabolas $y=x^{2}+3 x+6$ and $y=x^{2}+5 x+3$. Find the equation of their common tangent, as well as the points of tangency of this tangent with these parabolas.
1. Let the common tangent touch the first parabola at a point whose $x$-coordinate is $x_{1}$, and the second parabola at a point whose $x$-coordinate is $x_{2}$. The derivative of the equation of the first parabola at the point $x_{1}$ is $2 x_{1}+3$, which is the slope of the tangent at $x_{1}$, and similarly, by considering the second parabola, we get that the slope of the tangent is $2 x_{2}+5$. If the common tangent has the equation $y=k x+n$, then from the above we have $2 x_{1}+3=2 x_{2}+5=k$, and from there $x_{1}=x_{2}+1$. Further, since the points of tangency lie on both the tangent and the first and second parabolas, by considering the $y$-coordinates we get $\left(2 x_{1}+3\right) x_{1}+n=x_{1}^{2}+3 x_{1}+6$ and $\left(2 x_{2}+5\right) x_{2}+n=x_{2}^{2}+5 x_{2}+3$, from which it follows that
$n=6-x_{1}^{2}=3-x_{2}^{2}$, i.e., $x_{1}^{2}-x_{2}^{2}=3$. Substituting $x_{1}=x_{2}+1$ into this equation, we get $3=\left(x_{2}+1\right)^{2}-x_{2}^{2}=2 x_{2}+1$, and from there $x_{2}=1$, and then $x_{1}=x_{2}+1=2$. Further, we calculate $k=2 x_{1}+3=7$ and $n=6-x_{1}^{2}=2$, so the equation of the common tangent is $y=7 x+2$, and it touches the first parabola at the point $\left(x_{1}, 7 x_{1}+2\right)=(2,16)$, and the second at the point $\left(x_{2}, 7 x_{2}+2\right)=(1,9)$.