Решите ** отдельном листочке Кароче нужно решение

Question 1

Решите на отдельном листочке
Кароче нужно решение

Question 2

Решите задачу:

$(\frac{8a\sqrt{a}+b\sqrt{b}}{4\sqrt{a}+2\sqrt{b}}-\sqrt{ab})(\frac{4\sqrt{a}+2\sqrt{b}}{4a-b})^2=(\frac{(2\sqrt{a})^3+(\sqrt{b})^3}{2(2\sqrt{a}+\sqrt{b})}-\sqrt{ab})(\frac{2(2\sqrt{a}+\sqrt{b})}{(2\sqrt{a})^2-(\sqrt{b})^2})^2 =\\\\=\frac{(2\sqrt{a}+\sqrt{b})(4a-2\sqrt{ab}+b)-2\sqrt{ab}(2\sqrt{a}+\sqrt{b})}{2(2\sqrt{a}+\sqrt{b})}\cdot \frac{4(2\sqrt{a}+\sqrt{b})^2}{(2\sqrt{a}-\sqrt{b})^2(2\sqrt{a}+\sqrt{b})^2}=\\\\=(4a-2\sqrt{ab}+b-2\sqrt{ab})\cdot \frac{2}{(2\sqrt{a}-\sqrt{b})^2}=$

$=\frac{2(4a-4\sqrt{ab}+b)}{(2\sqrt{a}-\sqrt{b})^2}=\frac{2(2\sqrt{a}-\sqrt{b})^2}{(2\sqrt{a}-\sqrt{b})^2}=2$

$2)\; \frac{7\sqrt3\sqrt{a}-7\sqrt5\sqrt{b}}{6\sqrt3\sqrt{a}+6\sqrt5\sqrt{b}}:\frac{3a-5b}{9a+15b+6\sqrt{15ab}}=\frac{7(\sqrt{3a}-\sqrt{5b})}{6(\sqrt{3a}+\sqrt{5b})}\cdot \frac{3(3a+2\sqrt{15ab}+5b)}{(\sqrt{3a})^2-(\sqrt{5b})^2}=\\\\=\frac{7(\sqrt{3a}-\sqrt{5b})\cdot (\sqrt{3a}+\sqrt{5b})^2}{2(\sqrt{3a}+\sqrt{5b})(\sqrt{3a}-\sqrt{5b})(\sqrt{3a}+\sqrt{5b})}=\frac{7}{2}$

$3)\; (\frac{\sqrt{2a}-\sqrt{b}}{\sqrt{2a}+\sqrt{b}}-\frac{\sqrt{2a}+\sqrt{b}}{\sqrt{2a}-\sqrt{b}})(\sqrt{\frac{2b}{a}}-\sqrt{\frac{8a}{b}})=\\\\=\frac{(\sqrt{2a}-\sqrt{b})^2-(\sqrt{2a}+\sqrt{b})^2}{(\sqrt{2a}+\sqrt{b})(\sqrt{2a}-\sqrt{b})}\cdot \frac{\sqrt{2b}\sqrt{b}-\sqrt{8a}\sqrt{a}}{\sqrt{ab}}=\\\\=\frac{-4\sqrt{2ab}}{2a-b}\cdot \frac{\sqrt2b-2\sqrt2a}{\sqrt{ab}}=\frac{-4\sqrt2}{-(b-2a)}\cdot \frac{\sqrt2(b-2a)}{1}=4\cdot 2=8$