Question

ПОМОГИТЕ С 12 ПРИМЕРОМ,ДАЮ ОЧЕНЬ МНОГО БАЛЛОВ,ОЧЕНЬ НУЖНО!!!!!!!!!!!

Answer 1

$\Big (\frac{\sqrt[4]3-\sqrt[4]{27}}{3-\sqrt3}+\frac{1+3^{-0,5}}{3^{-0,25}}\Big )^2\cdot \Big (4-\frac{6}{\sqrt3}\Big )^{-\frac{1}{2}}=\\\\=\Big (\frac{\sqrt[4]3\cdot (1-\sqrt[4]{3^2})}{\sqrt3\cdot (\sqrt3-1)}+\frac{1+\frac{1}{\sqrt3}}{\frac{1}{\sqrt[4]3}}\Big )^2\cdot \Big (\frac{4\sqrt3-6}{\sqrt3}\Big )^{-\frac{1}{2}}=\\\\=\Big (\frac{1-\sqrt{3}}{\sqrt[4]3(\sqrt3-1)} +\frac{\sqrt[4]3\cdot (\sqrt3+1)}{\sqrt3} \Big )^2\cdot \Big (\frac{\sqrt3}{\sqrt3\cdot (4-2\sqrt3)}\Big )^{\frac{1}{2}}=$

$=\Big (-\frac{1}{\sqrt[4]3}+\frac{\sqrt3+1}{\sqrt[4]3}\Big )^2\cdot \frac{1}{\sqrt{4-2\sqrt3}} =\Big (\frac{\sqrt3}{\sqrt[4]3}\Big )^2\cdot \frac{1}{\sqrt{(1-\sqrt3)^2}}=\\\\=(\sqrt[4]3)^2\cdot \frac{1}{|1-\sqrt3|}=\frac{\sqrt3}{\sqrt3-1}\; ;\\\\\\\star \; \sqrt{4-2\sqrt3}=\sqrt{1+3-2\sqrt3}=\sqrt{1^2-2\cdot 1\cdot \sqrt3+(\sqrt3)^2}=\\\\=\sqrt{(1-\sqrt3)^2}=|\underbrace {1-\sqrt3}_{<0}|=-(1-\sqrt3)=\sqrt3-1\; \; \star$