Монгол Бодлогын Сан

Заавар: $\sqrt[3]{a^3}=a$, $\sqrt{a^2}=|a|$, $a^{-n}=\dfrac{1}{a^n}$ болохыг ашигла.

Бодолт: $$\left(\dfrac{\sqrt2}{2}\right)^{-6}=\left(\dfrac{2}{\sqrt2}\right)^6=(\sqrt2)^6=2^3$$ $$\left(\dfrac{\sqrt5}{5}\right)^{-2}=\left(\dfrac{5}{\sqrt5}\right)^2=(\sqrt5)^2$$ тул $$\sqrt[3]{\left(\dfrac{\sqrt2}{2}\right)^{-6}\cdot(2-\sqrt5)^3}+\sqrt{\left(\dfrac{\sqrt5}{5}\right)^{-2}\cdot(2-\sqrt5)^2}=$$ $$=\sqrt[3]{2^3\cdot(2-\sqrt5)^3}+\sqrt{(\sqrt5)^2\cdot(2-\sqrt5)^3}=$$ $$=2(2-\sqrt5)+|\sqrt5(2-\sqrt5)|=$$ $$=2(2-\sqrt5)+\sqrt5(\sqrt5-2)=$$ $$=4-2\sqrt5+5-2\sqrt5=9-4\sqrt5$$