Question

vă rog frumos! Repede ​

Answer 1

$\displaystyle \sqrt{6} \cdot \Bigg(\frac{4}{\sqrt{2} }-\frac{3}{\sqrt{3} }\Bigg)+\Bigg(\frac{6}{\sqrt{12} } +\frac{9}{\sqrt{27} }-\frac{16}{4\sqrt{8} }\Bigg)\cdot \frac{1}{2\sqrt{3}-\sqrt{2}}+\sqrt{18} =\\\\=\sqrt{6}\cdot \frac{4\sqrt{3} -3\sqrt{2} }{\sqrt{6} }+\Bigg(\frac{6}{2\sqrt{3} } +\frac{9}{3\sqrt{3} }-\frac{16}{4 \cdot 2\sqrt{2} }\Bigg) \cdot \frac{1}{2\sqrt{3}-\sqrt{2} }+3\sqrt{2} =$

$\displaystyle =4\sqrt{3} -3\sqrt{2} +\Bigg(\frac{6}{2\sqrt{3} } +\frac{9}{3\sqrt{3} }-\frac{16}{8\sqrt{2} } \Bigg) \cdot \frac{1}{2\sqrt{3} -\sqrt{2} } +3\sqrt{2} =\\\\=4\sqrt{3} -3\sqrt{2} +\frac{12 \cdot 6\cdot \sqrt{2} +8 \cdot 9\cdot \sqrt{2} -3 \cdot 16\cdot \sqrt{3} }{24\sqrt{6} } \cdot \frac{1}{2\sqrt{3}-\sqrt{2}} +3\sqrt{2}=\\\\= 4\sqrt{3} -3\sqrt{2} +\frac{72\sqrt{2}+ 72\sqrt{2}-48\sqrt{3} }{24\sqrt{6} } \cdot \frac{1}{2\sqrt{3}-\sqrt{2} } +3\sqrt{2} =$

$\displaystyle =4\sqrt{3} +\frac{144\sqrt{2}-48\sqrt{3} }{24\sqrt{6} \Big(2\sqrt{3} -\sqrt{2} \Big)} =4\sqrt{3} +\frac{48\Big(3\sqrt{2}-\sqrt{3} \Big) }{24\sqrt{6} \Big(2\sqrt{3} -\sqrt{2} \Big)} =\\\\=4\sqrt{3} +\frac{2\Big(3\sqrt{2} -\sqrt{3}\Big) }{\sqrt{6}\Big(2\sqrt{3} -\sqrt{2}\Big) }=4\sqrt{3} +\frac{6\sqrt{2} -2\sqrt{3} }{2\sqrt{18}-\sqrt{12} }=4\sqrt{3}+\frac{6\sqrt{2}-2\sqrt{3} }{2 \cdot 3\sqrt{2} -2\sqrt{3} } =\\\\ =4\sqrt{3} +\frac{6\sqrt{2}-2\sqrt{3} }{6\sqrt{2} -2\sqrt{3} } =4\sqrt{3}+1$