Номер 6 и 7,решите пожалуйста:))

Решите задачу:

$6)\; (3\sqrt{18}:\sqrt{27}+5\sqrt6-2):(\sqrt{54}-1)=\\\\=(\frac{3\cdot 3\sqrt{2}}{3\sqrt3}+5\sqrt{2\cdot 3} -2):(\sqrt{2\cdot 27}-1)=\\\\=(\sqrt3\cdot \sqrt2+5\sqrt2\cdot \sqrt3-2):(3\sqrt3\cdot \sqrt2-1)=\\\\=\frac{6\sqrt2\cdot \sqrt3-2}{3\sqrt2\cdot \sqrt3-1}=\frac{2(3\sqrt6-1)}{3\sqrt6-1}=2\\\\b)\; (2\sqrt{3,5}-\sqrt{126}+3)(3+\sqrt{56})=(2\sqrt{\frac{7}{2}}-\sqrt{9\cdot 14}+3)(3+\sqrt{7\cdot 8})=\\\\=(\sqrt{2\cdot 7}-3\sqrt{14}+3)(3+2\sqrt{7\cdot 2})=(3-2\sqrt{14})(3+2\sqrt{14})=\\\\=3^2-(2\sqrt{14})^2$ $=9-4\cdot 14=9-56=-47$

$7)\; \frac{\sqrt{7-4\sqrt3}}{\sqrt{2-\sqrt3}}\cdot \sqrt{2+\sqrt3}=\frac{\sqrt{(2-\sqrt3)^2}}{\sqrt{2-\sqrt3}}\cdot \sqrt{2+\sqrt3}=\\\\=\frac{|2-\sqrt3|}{\sqrt{2-\sqrt3}}\cdot \sqrt{2+\sqrt3}=\frac{2-\sqrt3}{\sqrt{2-\sqrt3}}\cdot \sqrt{2+\sqrt3}=\sqrt{2-\sqrt3}\cdot \sqrt{2+\sqrt3}=\\\\=\sqrt{(2-\sqrt3)(2+\sqrt3)}=\sqrt{2^2-(\sqrt3)^2}=\sqrt{4-3}=1$