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Question 1

Question 2

$1)\quad \frac{\sqrt[3]{243}}{\sqrt[3]{-9}} =\sqrt[3]{\frac{243}{-9}}=\sqrt[3]{-27}=\sqrt[3]{(-3)^3}=-3\\\\2)\quad \sqrt[4]{3\frac{3}{8}\cdot 1\frac{1}{2}}+ \frac{\sqrt[4]5}{\sqrt[4]{80}} =\sqrt[4]{ \frac{27}{8}\cdot \frac{3}{2}}+\sqrt[4]{\frac{5}{80}}=\sqrt[4]{ \frac{3^3\cdot 3}{2^3\cdot 2} }+\sqrt[4]{ \frac{1}{16} }=\\\\=\sqrt[4]{\frac{3^4}{2^4}}+\frac{1}{\sqrt[4]{2^4}}=\frac{3}{2}+\frac{1}{2}=\frac{4}{2}=2$

$3)\quad \sqrt[5]{-\frac{243}{1024}} \cdot \sqrt[3]{-4\frac{17}{27} }=\sqrt[5]{- \frac{3^5}{2^{10}}}\cdot \sqrt[3]{-\frac{125}{27} }=-\frac{3}{2^5}\cdot \left (\sqrt[3]{-\frac{5^3}{3^3}}\right )=\\\\=-\frac{3}{32}\cdot (-\frac{5}{3})=\frac{5}{32}\\\\4)\quad \sqrt[5]{160\cdot 625}=\sqrt[5]{32\cdot 5\cdot 5^4}=\sqrt[5]{2^5\cdot 5^5}=2\cdot 5=10\\\\5)\quad (12\sqrt{45}-6\sqrt{20}):3\sqrt5= \frac{12\sqrt{45}}{3\sqrt5}-\frac{6\sqrt{20}}{3\sqrt5} =4\sqrt{9}-2\sqrt4=\\\\=4\cdot 3-2\cdot 2=12-4=8$

8\\\\\sqrt{(x-10)^2}+\sqrt{(x-8)^2} =|x-10|+|x-8|=\\\\=|\sqrt{73}-10|+|\sqrt{73}-8|=(10-\sqrt{73})+(\sqrt{73}-8)=10-8=2\\\\7)\quad 32\cdot \left (\frac{\sqrt{x}+7}{\sqrt{x}-7}+\frac{\sqrt{x}-7}{\sqrt{x}+7}-\frac{196}{x-49}\right )^{-3}=" alt="6)\quad x=\sqrt{73}\approx 8,54\\\\\sqrt{73}<10\; ,\; \; \sqrt{73}>8\\\\\sqrt{(x-10)^2}+\sqrt{(x-8)^2} =|x-10|+|x-8|=\\\\=|\sqrt{73}-10|+|\sqrt{73}-8|=(10-\sqrt{73})+(\sqrt{73}-8)=10-8=2\\\\7)\quad 32\cdot \left (\frac{\sqrt{x}+7}{\sqrt{x}-7}+\frac{\sqrt{x}-7}{\sqrt{x}+7}-\frac{196}{x-49}\right )^{-3}=" align="absmiddle" class="latex-formula">

$=32\cdot \left ( \frac{(\sqrt{x}+7)^2+(\sqrt{x}-7)^2-196}{(\sqrt{x}-7)(\sqrt{x}+7)} \right )^{-3}=$

$=32\cdot \left ( \frac{x+14\sqrt{x}+49+x-14\sqrt{x}+49-196}{(\sqrt{x}-7)(\sqrt{x}+7)} \right )^{-3}=32\cdot \left (\frac{2x-98}{x-49}\right )^{-3}=$

$=32\cdot \left ( \frac{2(x-49)}{x-49} \right )^{-3}=32\cdot 2^{-3}=32\cdot \frac{1}{8}=4$

NNNLLL54_zn · Answer 1 · 2018-05-03T20:55:13+0000

$1)\quad \frac{\sqrt[3]{243}}{\sqrt[3]{-9}} =\sqrt[3]{\frac{243}{-9}}=\sqrt[3]{-27}=\sqrt[3]{(-3)^3}=-3\\\\2)\quad \sqrt[4]{3\frac{3}{8}\cdot 1\frac{1}{2}}+ \frac{\sqrt[4]5}{\sqrt[4]{80}} =\sqrt[4]{ \frac{27}{8}\cdot \frac{3}{2}}+\sqrt[4]{\frac{5}{80}}=\sqrt[4]{ \frac{3^3\cdot 3}{2^3\cdot 2} }+\sqrt[4]{ \frac{1}{16} }=\\\\=\sqrt[4]{\frac{3^4}{2^4}}+\frac{1}{\sqrt[4]{2^4}}=\frac{3}{2}+\frac{1}{2}=\frac{4}{2}=2$

$3)\quad \sqrt[5]{-\frac{243}{1024}} \cdot \sqrt[3]{-4\frac{17}{27} }=\sqrt[5]{- \frac{3^5}{2^{10}}}\cdot \sqrt[3]{-\frac{125}{27} }=-\frac{3}{2^5}\cdot \left (\sqrt[3]{-\frac{5^3}{3^3}}\right )=\\\\=-\frac{3}{32}\cdot (-\frac{5}{3})=\frac{5}{32}\\\\4)\quad \sqrt[5]{160\cdot 625}=\sqrt[5]{32\cdot 5\cdot 5^4}=\sqrt[5]{2^5\cdot 5^5}=2\cdot 5=10\\\\5)\quad (12\sqrt{45}-6\sqrt{20}):3\sqrt5= \frac{12\sqrt{45}}{3\sqrt5}-\frac{6\sqrt{20}}{3\sqrt5} =4\sqrt{9}-2\sqrt4=\\\\=4\cdot 3-2\cdot 2=12-4=8$

8\\\\\sqrt{(x-10)^2}+\sqrt{(x-8)^2} =|x-10|+|x-8|=\\\\=|\sqrt{73}-10|+|\sqrt{73}-8|=(10-\sqrt{73})+(\sqrt{73}-8)=10-8=2\\\\7)\quad 32\cdot \left (\frac{\sqrt{x}+7}{\sqrt{x}-7}+\frac{\sqrt{x}-7}{\sqrt{x}+7}-\frac{196}{x-49}\right )^{-3}=" alt="6)\quad x=\sqrt{73}\approx 8,54\\\\\sqrt{73}<10\; ,\; \; \sqrt{73}>8\\\\\sqrt{(x-10)^2}+\sqrt{(x-8)^2} =|x-10|+|x-8|=\\\\=|\sqrt{73}-10|+|\sqrt{73}-8|=(10-\sqrt{73})+(\sqrt{73}-8)=10-8=2\\\\7)\quad 32\cdot \left (\frac{\sqrt{x}+7}{\sqrt{x}-7}+\frac{\sqrt{x}-7}{\sqrt{x}+7}-\frac{196}{x-49}\right )^{-3}=" align="absmiddle" class="latex-formula">

$=32\cdot \left ( \frac{(\sqrt{x}+7)^2+(\sqrt{x}-7)^2-196}{(\sqrt{x}-7)(\sqrt{x}+7)} \right )^{-3}=$

$=32\cdot \left ( \frac{x+14\sqrt{x}+49+x-14\sqrt{x}+49-196}{(\sqrt{x}-7)(\sqrt{x}+7)} \right )^{-3}=32\cdot \left (\frac{2x-98}{x-49}\right )^{-3}=$

$=32\cdot \left ( \frac{2(x-49)}{x-49} \right )^{-3}=32\cdot 2^{-3}=32\cdot \frac{1}{8}=4$