Вычисления и преобразования

$$\sqrt{\sqrt{2013 \cdot 2015 \cdot 2019 \cdot 2021+36}+10}=\sqrt{\sqrt{(2017-4) \cdot (2017-2) \cdot (2017+2) \cdot (2017+4)+36}+10}=$$

$$=\sqrt{\sqrt{(2017^2-16) \cdot (2017^2-4) +36}+10}=\sqrt{\sqrt{(2017^4-16 \cdot 2017^2-4 \cdot 2017^2 +64+36}+10}=$$

$$=\sqrt{\sqrt{(2017^4-20 \cdot 2017^2+100}+10}=\sqrt{\sqrt{(2017^2-10)^2}+10}=\sqrt{2017^2-10+10}=2017$$

Ответ: 2017

Домножим $a$ на $2-\sqrt[8]{240}$ - и числитель, и знаменатель.

$$a=\frac{8(2-\sqrt[8]{240})}{(16+\sqrt{240})(4+\sqrt[4]{240})(2+\sqrt[8]{240}(2-\sqrt[8]{240})}=\frac{8(2-\sqrt[8]{240})}{(16+\sqrt{240})(4+\sqrt[4]{240})(4-\sqrt[4]{240})}=$$

$$=\frac{8(2-\sqrt[8]{240})}{(16+\sqrt{240})(16-\sqrt{240})}=\frac{8(2-\sqrt[8]{240})}{(256-240)}= \frac{8(2-\sqrt[8]{240})}{16}= \frac{(2-\sqrt[8]{240})}{2}$$

$$2a=2-\sqrt[8]{240}$$

$$2-2a =\sqrt[8]{240}$$

$$1-a=\frac{\sqrt[8]{240}}{2}$$

$$(1-a)^8=\frac{240}{256}=\frac{15}{16}$$

$$1-(1-a)^8=\frac{1}{16}$$

Тогда искомое

$$\frac{1}{1-(1-a)^8}=16$$

Ответ: 16

$$\left(\cos 36^{\circ}+\cos108^{\circ}\right)\cdot \sin 10^{\circ}\cdot \sin 50^{\circ}\cdot \sin 70^{\circ}=\left(2\cos 36^{\circ}\cos72^{\circ}\right)\cdot \sin 10^{\circ}\cdot \sin 50^{\circ}\cdot \sin 70^{\circ}=$$

$$=\frac{\left(2\cos 36^{\circ}\cos72^{\circ}\right) \sin 36^{\circ}}{\sin 36^{\circ}}\cdot \sin 10^{\circ}\cdot \sin 50^{\circ}\cdot \sin 70^{\circ}=\frac{\left(\cos 72^{\circ}\sin72^{\circ}\right) }{\sin 36^{\circ}}\cdot \sin 10^{\circ}\cdot \sin 50^{\circ}\cdot \sin 70^{\circ}=$$

$$=\frac{\sin144^{\circ}}{2\sin 36^{\circ}}\cdot \sin 10^{\circ}\cdot \sin 50^{\circ}\cdot \sin 70^{\circ}=\frac{\sin(180^{\circ}-36^{\circ})}{2\sin 36^{\circ}}\cdot \sin 10^{\circ}\cdot \sin 50^{\circ}\cdot \sin 70^{\circ}=$$

$$=\frac{\sin36^{\circ}}{2\sin 36^{\circ}}\cdot \sin 10^{\circ}\cdot \sin 50^{\circ}\cdot \sin 70^{\circ}=\frac{1}{2}\cdot \cos 80^{\circ}\cdot \cos 40^{\circ}\cdot \cos 20^{\circ}=$$

$$=\frac{1}{2}\cdot \cos 80^{\circ}\cdot \cos 40^{\circ}\cdot\frac{ \cos 20^{\circ}\cdot 2\sin 20^{\circ}}{2\sin 20^{\circ}}=\frac{1}{2}\cdot \cos 80^{\circ}\cdot \cos 40^{\circ}\cdot\frac{ \sin 40^{\circ}}{2\sin 20^{\circ}}=$$

$$=\frac{1}{2}\cdot \cos 80^{\circ}\cdot\frac{ \sin 80^{\circ}}{4\sin 20^{\circ}}=\frac{1}{2}\cdot \frac{ \sin 160^{\circ}}{8\sin 20^{\circ}}=\frac{1}{2}\cdot \frac{ \sin (180^{\circ}-20^{\circ})}{8\sin 20^{\circ}}=\frac{1}{16}$$

Ответ: $\frac{1}{16}$