Помогите пожалуйста с A5 B1 B2

Решите задачу:

$A5)\frac{b}{\sqrt{7}} = \frac{b*\sqrt{7}} {\sqrt{7}*\sqrt{7}}= \frac{b\sqrt{7}} {7}\\\\\frac{5}{\sqrt{13}+\sqrt{3}}= \frac{5*(\sqrt{13}-\sqrt{3})} {(\sqrt{13}+\sqrt{3})(\sqrt{13}-\sqrt{3})}= \frac{5(\sqrt{13}-\sqrt{3})}{13-3}= \frac{5(\sqrt{13}-\sqrt{3})} {10}= \frac{\sqrt{13}-\sqrt{3}} {2}$

$B1)\sqrt{(2-\sqrt{7})^{2}} }+ \sqrt{(3-\sqrt{7})^{2}}=|2- \sqrt{7}|+|3- \sqrt{7}|= \sqrt{7}-2+3- \sqrt{7}=1\\\\B2)\frac{\sqrt{5}-\sqrt{3}} {\sqrt{5}+\sqrt{3}}- \frac{\sqrt{5}+\sqrt{3}} {\sqrt{5}-\sqrt{3}}= \frac{(\sqrt{5}-\sqrt{3})^{2}-(\sqrt{5}+\sqrt{3})^{2}} {(\sqrt{5}+\sqrt{3})(\sqrt{5}-\sqrt{3})}=\frac{5-2\sqrt{15}+3-5-2\sqrt{15}-3}{5-3}=- \frac{4\sqrt{15}} {2} =-2 \sqrt{15}$