Алгебра 1 простое задание 100 баллов

$\sqrt{4\pm\frac{\sqrt{15}}{2}}=\sqrt{\frac{16\pm 2\sqrt{15}}{4}}= \frac{\sqrt{(\sqrt{15}\pm 1)^2}}{2}=\frac{|\sqrt{15}\pm 1|}{2}=\frac{\sqrt{15}\pm 1}{2}$

$f(\frac{\sqrt{15}}{2})=\frac{2\sqrt{15}+\frac{15}{4}}{\frac{5+\sqrt{15}}{2}}- \frac{2\sqrt{15}-\frac{15}{4}}{\frac{5-\sqrt{15}}{2}}= \frac{8\sqrt{15}+15}{2(5+\sqrt{15})}-\frac{8\sqrt{15}-15}{2(5-\sqrt{15})}=$

$=\frac{\sqrt{15}}{2}\cdot\frac{((8+\sqrt{15})(5-\sqrt{15})-(8-\sqrt{15})(5+\sqrt{15})}{(5+\sqrt{15})(5-\sqrt{15})}=\frac{\sqrt{15}}{2}\cdot\frac{-6\sqrt{15}}{10}= -\frac{9}{2}$

Ответ: $-\frac{9}{2}$