Answer :
Answer:
- 11 = 103
- 12 = 1.732
Step-by-step explanation:
[tex]\frac{1}{216^{\frac{-2}{3}}}+\frac{1}{256^{-\frac{3}{4}}}+\frac{1}{243^{-\frac{1}{5}}}\\\\Let \: start \:with ; \: \frac{1}{216^{\frac{-2}{3}}}\\\\\mathrm{Apply\:the\:fraction\:rule}:\\\quad \frac{-a}{b}=-\frac{a}{b}\\=\frac{1}{216^{-\frac{2}{3}}}\\216^{-\frac{2}{3}}\\\\\mathrm{Apply\:exponent\:rule}:\quad \:a^{-b}=\frac{1}{a^b}\\=\frac{1}{216^{\frac{2}{3}}}\\\\216^{\frac{2}{3}}\\=\left(6^3\right)^{\frac{2}{3}}\\\left(6^3\right)^{\frac{2}{3}}=6^{3\times \frac{2}{3}}=6^2\\=36=\frac{1}{\frac{1}{36}}\\[/tex]
[tex]\mathrm{Apply\:the\:fraction\:rule}:\quad \frac{1}{\frac{b}{c}}=\frac{c}{b}=\frac{36}{1}=36\\\frac{1}{256^{-\frac{3}{4}}}\\\\256^{-\frac{3}{4}}=\frac{1}{64}\\\\=\frac{1}{\frac{1}{64}} = 64/1 = 6\\\\\frac{1}{243^{-\frac{1}{5}}}\\\\243^{-\frac{1}{5}}=\frac{1}{3}\\\\=\frac{1}{\frac{1}{3}} = 3/1 = 3\\\\\\=36+64+3\\=103[/tex]
[tex]\frac{3+\sqrt{6}}{5\sqrt{3}-2\sqrt{12}-\sqrt{32}+\sqrt{50}}\\\\2\sqrt{12} = =4\sqrt{3}\\\\\sqrt{32}=4\sqrt{2}\\\\\sqrt{50}=5\sqrt{2}\\\\=\frac{3+\sqrt{6}}{5\sqrt{3}-4\sqrt{3}-4\sqrt{2}+5\sqrt{2}}\\\\5\sqrt{3}-4\sqrt{3}-4\sqrt{2}+5\sqrt{2}=\sqrt{3}+\sqrt{2}\\\\\mathrm{Add\:similar\:elements:}\:-4\sqrt{2}+5\sqrt{2}=\sqrt{2}\\\\=5\sqrt{3}-4\sqrt{3}+\sqrt{2}\\\\\mathrm{Add\:similar\:elements:}\:5\sqrt{3}-4\sqrt{3}=\sqrt{3}\\\\[/tex]
[tex]\mathrm{Add\:similar\:elements:}\:5\sqrt{3}-4\sqrt{3}=\sqrt{3}\\\\=\sqrt{3}+\sqrt{2}\\\=\frac{3+\sqrt{6}}{\sqrt{3}+\sqrt{2}}\\\mathrm{Multiply\:by\:the\:conjugate}\:\frac{\sqrt{3}-\sqrt{2}}{\sqrt{3}-\sqrt{2}}\\\\=\frac{\left(3+\sqrt{6}\right)\left(\sqrt{3}-\sqrt{2}\right)}\\{\left(\sqrt{3}+\sqrt{2}\right)\left(\sqrt{3}-\sqrt{2}\right)}\\\\\left(3+\sqrt{6}\right)\left(\sqrt{3}-\sqrt{2}\right)=\sqrt{3}\\\\\left(\sqrt{3}+\sqrt{2}\right)\left(\sqrt{3}-\sqrt{2}\right)=1\\=\frac{\sqrt{3}}{1}\\\\=\sqrt{3}[/tex]