Сократитед дробь: A)3-b²/√3+b Б)5-√5/√10-√2 В)4x²+4x√a+a/2x+√a

Решите задачу:

$1) \ \frac{3-b^2}{ \sqrt{3}+b}=\frac{(\sqrt{3}-b)(\sqrt{3}+b)}{ \sqrt{3} +b}= \sqrt{3} -b \\\\ \\ 2) \ \frac{5- \sqrt{5} }{ \sqrt{10}- \sqrt{2} }= \frac{(5- \sqrt{5})(\sqrt{10}+ \sqrt{2}) }{ (\sqrt{10}- \sqrt{2})(\sqrt{10}+ \sqrt{2}) }= \frac{5 \sqrt{10}+5 \sqrt{2}- \sqrt{50}- \sqrt{10} }{ 10-2 }= \\\\ = \frac{4 \sqrt{10}+5 \sqrt{2}- 5\sqrt{2}}{ 8 }= \frac{4 \sqrt{10}}{ 8 }=\frac{\sqrt{10}}{2}$

$3) \ \frac{4x^2+4x \sqrt{a}+a }{2x+ \sqrt{a} }= \frac{(2x +\sqrt{a})^2}{2x+ \sqrt{a} }=2x+ \sqrt{a}$