Question

Номер 6 и 7,решите пожалуйста:))

Answer 1

$6); (3sqrt{18}:sqrt{27}+5sqrt6-2):(sqrt{54}-1)=\\=(frac{3cdot 3sqrt{2}}{3sqrt3}+5sqrt{2cdot 3} -2):(sqrt{2cdot 27}-1)=\\=(sqrt3cdot sqrt2+5sqrt2cdot sqrt3-2):(3sqrt3cdot sqrt2-1)=\\=frac{6sqrt2cdot sqrt3-2}{3sqrt2cdot sqrt3-1}=frac{2(3sqrt6-1)}{3sqrt6-1}=2\\b); (2sqrt{3,5}-sqrt{126}+3)(3+sqrt{56})=(2sqrt{frac{7}{2}}-sqrt{9cdot 14}+3)(3+sqrt{7cdot 8})=\\=(sqrt{2cdot 7}-3sqrt{14}+3)(3+2sqrt{7cdot 2})=(3-2sqrt{14})(3+2sqrt{14})=\\=3^2-(2sqrt{14})^2$$=9-4cdot 14=9-56=-47$

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