[tex]D)-\frac{7+7\sqrt{5})}{4}[/tex]
1) Examining that ratio, we can perform the following:
[tex]\begin{gathered} \frac{7}{1-\sqrt{5}} \\ \\ \frac{7\left(1+\sqrt{5}\right)}{\left(1-\sqrt{5}\right)\left(1+\sqrt{5}\right)} \\ \\ \frac{7+7\sqrt{5}}{1^2-(\sqrt{5})^2} \\ \\ \frac{7(1+\sqrt{5})}{-4} \\ \\ -\frac{7(1+\sqrt{5})}{4} \end{gathered}[/tex]
2) Note that when we multiply that ratio by their conjugates, that yields a difference between two squares. Note that on the top, there is the expanded version of this expression.
Thus, the answer is D
