In my research, I found two combinatorial identities. Mathematica can give the answers immediately, but I don't know how to prove them. Could someone help me? Thank you!

Here are they:

Let $\alpha$, $\beta$ be two arbitrary complex numbers, and $k$, $l$ be two positive integers, then:

\begin{align*} \sum_{m=1}^k m \binom{\frac{\alpha+\beta}{\beta}k}{k-m}\binom{\frac{\alpha+\beta}{\beta}l}{l+m} =&\frac{\alpha\,k\,l}{(k+l)(\alpha+\beta)}\binom{\frac{\alpha+\beta}{\beta}k}{k}\binom{\frac{\alpha+\beta}{\beta}l}{l}\\ \sum_{m=1}^k m \binom{\frac{\alpha+\beta}{\beta}k}{k-m}\binom{\frac{\alpha+\beta}{\alpha}l}{l-m} =&\frac{\alpha\,\beta\,k\,l}{(k\,\alpha+l\,\beta)(\alpha+\beta)}\binom{\frac{\alpha+\beta}{\beta}k}{k}\binom{\frac{\alpha+\beta}{\alpha}l}{l} \end{align*}

Actually, the positive integer $l$ can be also a complex number. Then, if one replaces $\frac{\alpha}{\beta}l$ by $\tilde{l}$ in the first identity, it becomes equivalent to the second identity. Besides this observation, I can only prove some trivial cases.