(√27-2)(2-3√3) (5√3+√27):√3 (√5-1)^2-(2√5+1)^2

$(\sqrt{27}-2)(2-3\sqrt{3})=(3\sqrt{3}-2)(2-3\sqrt{3})=- (2-3\sqrt{3})(2-3\sqrt{3})=-(2-3\sqrt{3})^2=\\\ =-(4-12\sqrt{3}+27)=-(31-12\sqrt{3})=12\sqrt{3}-31$

$\frac{5\sqrt{3}+\sqrt{27}}{\sqrt{3}}=\frac{5\sqrt{3}+3\sqrt{3}}{\sqrt{3}}=\frac{8\sqrt{3}}{\sqrt{3}}=8$

$(\sqrt{5}-1)^2-(2\sqrt{5}+1)^2=((\sqrt{5}-1)-(2\sqrt{5}+1))((\sqrt{5}-1)+(2\sqrt{5}+1))=\\\ =(\sqrt{5}-1-2\sqrt{5}-1))(\sqrt{5}-1+2\sqrt{5}+1)=(-\sqrt{5}-2)*3\sqrt{5}=-15-6\sqrt{5}$
можно и по другому
$(\sqrt{5}-1)^2-(2\sqrt{5}+1)^2=5-2\sqrt{5}+1-(20+4\sqrt{5}+1)=5-2\sqrt{5}+1-20-4\sqrt{5}-1=\\\ =-15-6\sqrt{5}$