jmc

The numerator $x^2-80x+1551$ factors as $(x-47)(x-33)$, and the denominator $x^2+57x-2970$ factors as $(x-33)(x+90)$, so $$\frac{x^2-80x+1551}{x^2+57x-2970}=\frac{(x-47)(x-33)}{(x-33)(x+90)}=\frac{x-47}{x+90}.$$Then $\alpha =47$ and $\beta =90$, so $\alpha +\beta =47+90=\boxed{137}$.

We can also solve the problem using Vieta's formulas, which states that the sum of the roots of the quadratic $a{x}^2+bx+c=0$ is $-b/a$. The only way that the right-hand side $\frac{x^2-80x+1551}{x^2+57x-2970}$ can simplify to the left-hand side $\frac{x-\alpha }{x+\beta }$ is if $x^2-80x+1551$ and $x^2+57x-2970$ have a root in common. Call this common root $\gamma$.

Then the roots of $x^2-80x+1551=0$ are $\alpha$ and $\gamma$, so $\alpha +\gamma =80$. Similarly, the roots of $x^2+57x-2970=0$ are $-\beta$ and $\gamma$, so $-\beta +\gamma =-57$. Subtracting these equations, we get $\alpha +\beta =80-(-57)=\boxed{137}$.