Помогите упростить пожалуйста (подробно)

Question 1

Question 2

Упростить

$\left (\frac{1}{a+\sqrt{2}}-\frac{a^2+2}{a^3+2\sqrt{2}} \right )^{-1} \cdot \left(\frac{a}{2}-\frac{1}{\sqrt{2}}+\frac{1}{a} \right )^{-1} \cdot \frac{\sqrt{2}}{a+\sqrt{2}}$

Решение

$a^3+2\sqrt{2}=a^3+(\sqrt{2})^3=(a+\sqrt{2})(a^2-\sqrt{2}a+2)$

$\frac{1}{a+\sqrt{2}}-\frac{a^2+2}{a^3+2\sqrt{2}}=\frac{1}{a+\sqrt{2}}-\frac{a^2+2}{(a+\sqrt{2})(a^2-\sqrt{2}a+2)}=\frac{a^2-\sqrt{2}a+2-a^2-2}{(a+\sqrt{2})(a^2-\sqrt{2}a+2)}=\frac{-\sqrt{2}a}{(a+\sqrt{2})(a^2-\sqrt{2}a+2)}$

$\left (\frac{1}{a+\sqrt{2}}-\frac{a^2+2}{a^3+2\sqrt{2}} \right )^{-1}=\left (\frac{-\sqrt{2}a}{(a+\sqrt{2})(a^2-\sqrt{2}a+2)} \right )^{-1}=-\frac{(a+\sqrt{2})(a^2-\sqrt{2}a+2)}{\sqrt{2}a}$

$\left (\frac{a}{2}-\frac{1}{\sqrt{2}}+\frac{1}{a} \right )^{-1}=\left (\frac{a^2-\sqrt{2}a+2}{2a} \right )^{-1}=\frac{2a}{a^2-\sqrt{2}a+2}$

$\left (\frac{1}{a+\sqrt{2}}-\frac{a^2+2}{a^3+2\sqrt{2}} \right )^{-1} \cdot \left(\frac{a}{2}-\frac{1}{\sqrt{2}}+\frac{1}{a} \right )^{-1}=-\frac{(a+\sqrt{2})(a^2-\sqrt{2}a+2)}{\sqrt{2}a} \cdot \frac{2a}{a^2-\sqrt{2}a+2}=-\sqrt{2}(a+\sqrt{2})$

$\left (\frac{1}{a+\sqrt{2}}-\frac{a^2+2}{a^3+2\sqrt{2}} \right )^{-1} \cdot \left(\frac{a}{2}-\frac{1}{\sqrt{2}}+\frac{1}{a} \right )^{-1}\cdot\frac{\sqrt{2}}{a+\sqrt{2}}=-\sqrt{2}(a+\sqrt{2})\cdot\frac{\sqrt{2}}{a+\sqrt{2}}=-2$

Ответ: -2

Question 3

Спасибо))