Question

Памагите !!!! Пожалуйста !!!!

Answer 1

$1)\; \; \sqrt[3]{5-\sqrt{17}}\cdot \sqrt[3]{5+\sqrt{17}}=\sqrt[3]{(5-\sqrt{17})(5+\sqrt{17})}=\sqrt[3]{25-17}=\sqrt[3]{8}=2\\\\2)\; \; \sqrt[4]{26+\sqrt{51}}\cdot \sqrt[4]{26-\sqrt{51}}=\sqrt[4]{(26+\sqrt{51})(26-\sqrt{51})}=\sqrt[4]{676-51}=\\\\=\sqrt[4]{625}=\sqrt[4]{5^4}=5\\\\3)\; \; \sqrt[3]{343\, m^6\, n^9}=\sqrt[3]{7^3\, (m^2)^3\, (n^3)^3}=7\, m^2\, n^3$

$4)\; \; y\geq 0\; ,\; \; z\leq 0\\\\\sqrt[4]{16\, x^8\, y^4\, z^{12}}=\sqrt[4]{2^4\, (x^2)^4\, y^4\, (z^3)^4}=2\cdot |\underbrace {x^2}_{\geq 0}|\cdot |\underbrace {y}_{\geq 0}|\cdot |\underbrace {z^3}_{\leq 0}|=2\, x^2\cdot y\cdot (-z^3)=\\\\=-2\, x^2\, y\, z^3\\\\\\5)\; \; x\leq 0\\\\3x\sqrt[8]{256\, x^{24}}=3x\sqrt[8]{2^8\, (x^3)^8}=3x\cdot 2\cdot |\underbrace {x^3}_{\leq 0}|=6x\cdot (-x^3)=-6x^4$

$6)\; \; a\geq 23\; \; \Rightarrow \; \; \; (a-23)\geq 0\\\\\sqrt[6]{(a-23)^6}=|\underbrace {a-23}_{\geq 0}|=a-23\\\\\\7)\; \; y\leq -3\; \; \Rightarrow \; \; \; (y+3)\leq 0\\\\\sqrt[8]{(y+3)^8}=|\underbrace {y+3}_{\leq 0}|=-(y+3)=-y-3$