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Ответ:

Применяем свойства корней:  [tex]\boldsymbol{\sqrt[n]{a^{k}}=\sqrt[n\cdot p]{a^{k\cdot p}}\ \ ,\ \ \sqrt[n]{a}\cdot \sqrt[n]{b}=\sqrt[n]{a\cdot b}}[/tex]  .  

[tex]\displaystyle (\sqrt{27}-2)(2-3\sqrt3)=(\sqrt{27}-2)(2-\sqrt{27})=-(2-\sqrt{27})^2=-(4-4\sqrt{27}+27)=\\\\=4\sqrt{27}-31=4\cdot 3\sqrt3-31=12\sqrt3-31\\\\\\(\sqrt{50}+4\sqrt2)\cdot \sqrt2=\sqrt{100}+4\cdot 2=10+8=18\\\\\sqrt{20}:\sqrt{5}=\dfrac{\sqrt{5}\cdot \sqrt4}{\sqrt5}=\sqrt4=2\\\\\\\sqrt{12}:\sqrt{27}=\dfrac{\sqrt{4\cdot 3}}{\sqrt{9\cdot 3}}=\dfrac{2\sqrt3}{3\sqrt3}=\dfrac{2}{3}[/tex]

[tex]\sqrt[4]{512\cdot 216}=\sqrt[4]{2^9\cdot 6^3}=\sqrt[4]{2^9\cdot (2\cdot 3)^3}=\sqrt[4]{2^{12}\cdot 3^3}=2^3\sqrt[4]{3^3}=8\sqrt[4]{27}[/tex]

[tex]\sqrt[4]{324}\cdot \sqrt[4]{4}=\sqrt[4]{2^2\cdot 3^4}\cdot \sqrt[4]{2^2}=\sqrt[4]{2^2\cdot 3^4\cdot 2^2}=\sqrt[4]{2^4\cdot 3^4}=\sqrt[4]{6^4}=6[/tex]

[tex]\sqrt[4]{\sqrt[3]{25}}\cdot \sqrt[6]{5^5}=\sqrt[12]{25}\cdot \sqrt[12]{(5^5)^2}=\sqrt[12]{5^2\cdot 5^{10}}=\sqrt[12]{5^{12}}=5\\\\\\\dfrac{\sqrt3\cdot \sqrt[3]{9}}{\sqrt[6]{3}}=\dfrac{\sqrt[6]{3^3}\cdot \sqrt[6]{9^2}}{\sqrt[6]{3}}=\dfrac{\sqrt[6]{3^3\cdot 3^4}}{\sqrt[4]{3}}=\dfrac{\sqrt[6]{3^{7}}}{\sqrt[6]{3}}=\sqrt[6]{\dfrac{3^{7}}{3}}=\sqrt[6]{3^6}=3[/tex]