Помогите пожалуйста, срочно 11класс

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[latex]1)\; \; \; x^3-125=0\\\\x^3=125\\\\x^3=5^3\\\\x=5\\\\2)\; \; (\sqrt[6]{x})^2-3\cdot\sqrt[6]{x}=4\\\\t=\sqrt[6]{x} \geq 0\; ,\; \; \; t^2-3t-4=0\\\\t_1=-1,\; \; t_2=4\; \; (teorema\; Vieta)\\\\t \geq 0\; \; \Rightarrow \; \; \; \sqrt[6]{x}=4\; \; \; \Rightarrow \; \; x=4^6=4096\\\\3)\; \; \; \sqrt[3]{12+4\sqrt5} \cdot \sqrt[3]{12-4\sqrt5}=\sqrt[3]{(12+4\sqrt5)(12-4\sqrt5)}=\\\\=\sqrt[3]{12^2-16\cdot 5} = \sqrt[3]{64} =\sqrt[3]{4^3}=4\\\\4)\; \; a\ \textless \ 0\; ,\\\\\sqrt[4]{a^4}+\sqrt[3]{a^3}=|a|+a=-a+a=0[/latex] [latex]5)\; \; \frac{3-\sqrt3}{3+\sqrt3}=\frac{(3-\sqrt3)(3-\sqrt3)}{(3+\sqrt3)(3-\sqrt3)}=\frac{12-6\sqrt3}{9-3}=\frac{6(2-\sqrt3)}{6}=2-\sqrt3\\\\6)\; \; \frac{\sqrt[8]{x^3}-\sqrt[4]{y^3}}{ \sqrt[4]{x}+\sqrt{y}+\sqrt[8]{xy^2} }=\frac{(\sqrt[8]{x})^3-(\sqrt[4]{y})^3}{ \sqrt[4]{x}+\sqrt{y}+\sqrt[8]{xy^2} }=\frac{(\sqrt[8]{x}-\sqrt[4]{y})(\sqrt[8]{x^2}+\sqrt[8]{x}\sqrt[4]{y}+\sqrt[4]{y^2})}{\sqrt[4]{x}+\sqrt{y}+\sqrt[8]{xy^2}}=[/latex] [latex]= \frac{(\sqrt[8]{x}-\sqrt[4]{y})(\sqrt[4]{x}+\sqrt[8]{xy^2}+\sqrt{y})}{\sqrt[4]{x}+\sqrt{y}+\sqrt[8]{xy^2}}=\sqrt[8]{x}-\sqrt[4]{y}[/latex] [latex]7)\; \; \sqrt{29-12\sqrt5}-\sqrt{29+12\sqrt5}=\\\\=\sqrt{9+20-2\cdot 3\cdot 2\sqrt5}-\sqrt{9+20+2\cdot 3\cdot 2\sqrt5}=\\\\=\sqrt{(3-2\sqrt5)^2}-\sqrt{(3+2\sqrt5)^2}=|\underbrace {3-2\sqrt5}_{\ \textless \ 0}|-|\underbrace {3+2\sqrt5}_{\ \textgreater \ 0}|=\\\\=(2\sqrt5-3)-(3+2\sqrt5)=-6[/latex]