Помогите с 2до 8номера дам 25б

Question 1

Question 2

Решите задачу:

$2)\; \; \sqrt[3]{2\sqrt3-4}\cdot \sqrt[3]{2\sqrt3+4}=\sqrt[3]{4\cdot 3-16}=\sqrt[3]{-2}=-\sqrt[3]2\approx -1,26\\\\-\sqrt[3]{2}\in [-2,-1]\\\\3)\; \; 5ab^2\sqrt[3]{a^2b}=\sqrt[3]{(5ab^2)^3\cdot a^2b}=\sqrt[3]{125a^3b^6\cdot a^2b}=\sqrt[3]{125a^5b^7}\\\\4)\; \; y\sqrt[10]{y\sqrt[3]{y^2}}:\sqrt[6]{y^5}=y\sqrt[10]{\sqrt[3]{y^3\cdot y^2}}:\sqrt[6]{y^5}=\\\\=y\sqrt[10]{\sqrt[3]{y^5}}:\sqrt[6]{y^5}=y\sqrt[30]{y^5}:\sqrt[6]{y^5}=y\cdot \sqrt[6]{y}:\sqrt[6]{y^5}=$

$= \frac{y\sqrt[6]{y}}{\sqrt[6]{y^5}} = \sqrt[6]{\frac{y^7}{y^5} }=\sqrt[6]{y^2}=\sqrt[3]{y}$

$5)\; \; a= \frac{9}{16} ;\\\frac{a+\sqrt{a}}{1+\sqrt{a}} + \frac{a-1}{1+\sqrt{a} } = \frac{2a+\sqrt{a}-1}{1+\sqrt{a}} = \frac{\frac{18}{16}+\sqrt{\frac{9}{16}}-1}{1+\sqrt{\frac{9}{16}}} = \frac{\frac{18}{16}+\frac{3}{4}-1}{1+\frac{3}{4}} = \frac{\frac{14}{16}}{\frac{7}{4}} = \\\\=\frac{14\cdot 4}{16\cdot 7} =\frac{2}{4}=\frac{1}{2}\\\\6)\; \; \frac{\sqrt{\sqrt{c}+\sqrt{d}}\cdot \sqrt{\sqrt{c}-\sqrt{d}}}{d-c} =\frac{\sqrt{c-d}}{-(c-d)} = -\frac{1}{\sqrt{c-d}}$

$7)\; \; \sqrt{2m\sqrt[3]{\frac{1}{4m^2}\sqrt{\frac{n}{m}}}}:\sqrt[12]{nm}=\sqrt{\sqrt[3]{\frac{8m^3}{4m^2}\sqrt{\frac{n}{m}}}}:\sqrt[12]{nm}=\\\\=\sqrt[6]{2m\sqrt{\frac{n}{m}}}:\sqrt[12]{nm}=\sqrt[6]{\sqrt{4m^2\cdot \frac{n}{m}}}:\sqrt[12]{nm}=\\\\=\sqrt[12]{4mn}\cdot \frac{1}{\sqrt[12]{nm}}=\sqrt[12]{\frac{4mn}{nm}}=\sqrt[12]{4}$

$8)\; \; t=3;\\\\\frac{\sqrt{t^3-8}}{\sqrt{t-2}} = \frac{\sqrt{(t-2)(t^2+2t+4)}}{\sqrt{t-2}} =\sqrt{t^2+2t+4}=\sqrt{9+6+4}=\\\\=\sqrt{19}$