Пожалуйста помогите решить, буду очень благодарен #318 (1-4)

[latex]1) \sqrt[3]{5}; \sqrt[4]{8}; \sqrt{3} \\HOK(3,4,2)=12\\( \sqrt[3]{5})^{12}=(5^{ \frac{1}{3})^{12}}=5^4=625\\( \sqrt[4]{8})^{12}=(8^{ \frac{1}{4})^{12}}=8^3=512\\( \sqrt{3})^{12}=(3^{ \frac{1}{2}})^{12}=3^4=81\\\\625\ \textgreater \ 512\ \textgreater \ 81\; \; =\ \textgreater \ \; \;\sqrt[3]{5}\ \textgreater \ \sqrt[4]{8}\ \textgreater \ \sqrt{3} \\\\\\2) \sqrt[4]{5}; \sqrt{2}; \sqrt[3]{3}\\OK(4;2;3)=12\\(\sqrt[4]{5})^{12}=(5^{ \frac{1}{4}})^{12}=5^3=125\\( \sqrt{2})^{12}=(2^{ \frac{1}{2}})^{12}=2^6=64\\( \sqrt[3]{3})^{12}=(3^{ \frac{1}{3}})^{12}=3^4=81[/latex] [latex]64\ \textless \ 81\ \textless \ 125\; \; =\ \textgreater \ \; \; \sqrt{2}\ \textless \ \sqrt[3]{3}\ \textless \ \sqrt[4]{5} [/latex] [latex]3) \sqrt{2* \sqrt[3]{2} }=2^{ \frac{1}{2}}*2^{ \frac{1}{6}}=2^{ \frac{4}{6}} = 2^{ \frac{2}{3}}\\ \sqrt[3]{2* \sqrt{2} }=2^{ \frac{1}{3}}*2^{ \frac{1}{6}}=2^{ \frac{3}{6}}=2^{ \frac{1}{2}}\\ \frac{2}{3} \ \textgreater \ \frac{1}{2}\; \; u\; \; 2\ \textgreater \ 1=\ \textgreater \ 2^{ \frac{2}{3}}\ \textgreater \ 2^{ \frac{1}{2}} \; \; =\ \textgreater \ \sqrt{2* \sqrt[3]{2} }\ \textgreater \ \sqrt[3]{2* \sqrt{2} }\\ [/latex] [latex]4) \sqrt[4]{8 \sqrt{2} }= 2^{ \frac{3}{4}}*2^{ \frac{1}{8}}=2^{ \frac{7}{8}}\\ \sqrt{2 \sqrt[4]{8} }=2^{ \frac{1}{2}}*(2^{ \frac{3}{4}})^{ \frac{1}{2}}=2^{ \frac{1}{2}}*2^{ \frac{3}{8}}=2^{ \frac{7}{8}}\\\\ \sqrt[4]{8 \sqrt{2} }= \sqrt{2 \sqrt[4]{8} } [/latex]