Сократите дробь:1) 2)помогите пожалуйста.срочно надо.

1)
$\frac{(\sqrt{3}-\sqrt{2})*\sqrt{72}}{3(2\sqrt{6}-\sqrt{16})(\sqrt[3]{64}+1)}=\\\\\frac{(\sqrt{3}-\sqrt{2})*\sqrt{2}*6}{3*\sqrt{2}*2(\sqrt{3}-\sqrt{2})*(4+1)}=\\\\\frac{2}{5}=0.4$
так как
$\sqrt{72}=\sqrt{2*36}=\sqrt{2}*\sqrt{36}=\sqrt{2}*6$
$\sqrt[3]{64}=\sqrt[3]{4^3}=4$
;
$2\sqrt{6}-\sqrt{16}=2\sqrt{3}*\sqrt{2}-\sqrt{4}*\sqrt{2}*\sqrt{2}=\\\\2*\sqrt{2}*(\sqrt{3}-\sqrt{2})$
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2)
$\frac{\sqrt{6}+\sqrt{3}-\sqrt{2}-1}{\sqrt{6}+2\sqrt{3}-\sqrt{2}-2}=\\\\\frac{\sqrt{3}*\sqrt{2}+\sqrt{3}*1-\sqrt{2}*1-1*1}{\sqrt{3}*\sqrt{2}+\sqrt{3}*2-1*\sqrt{2}-1*2}=\\\\\frac{\sqrt{3}(\sqrt{2}+1)-1*(\sqrt{2}+1)}{\sqrt{3}(\sqrt{2}+2)-1(\sqrt{2}+2)}=\\\\\frac{(\sqrt{3}-1)(\sqrt{2}+1)}{(\sqrt{3}-1)(\sqrt{2}+2)}=\frac{\sqrt{2}+1}{\sqrt{2}+2}$