Question

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отдаю 42 балла ​

Answer 1

$\displaystyle\bf\\1) \ \ \frac{3}{\sqrt{2} +\sqrt{3} } =\frac{3\cdot(\sqrt{2}-\sqrt{3} ) }{(\sqrt{2} +\sqrt{3} )\cdot(\sqrt{2} -\sqrt{3} )} =\frac{3\cdot(\sqrt{2}-\sqrt{3} ) }{(\sqrt{2} )^{2} -(\sqrt{3} )^{2} } =\\\\\=\frac{3\cdot(\sqrt{2}-\sqrt{3} ) }{2-3} =-3(\sqrt{2} -\sqrt{3})=3(\sqrt{3} -\sqrt{2} )\\\\\\2) \ \ \ \frac{\sqrt{11}-3 }{\sqrt{11} +3} =\frac{(\sqrt{11} -3)\cdot(\sqrt{11}-3) }{(\sqrt{11} +3)\cdot(\sqrt{11}-3) } =\frac{(\sqrt{11} -3)^{2} }{(\sqrt{11})^{2} -3^{2} }$

$\displaystyle\bf\\=\frac{3\cdot(\sqrt{2} -\sqrt{3}) }{2-3} =-3(\sqrt{2} -\sqrt{3} )=3\sqrt{3} -3\sqrt{2}$

$\displaystyle\bf\\\frac{\sqrt{11} -3}{\sqrt{11} +3} =\frac{(\sqrt{11} -3)(\sqrt{11} -3)}{(\sqrt{11} +3)(\sqrt{11} -3)} =\frac{(\sqrt{11}-3)^{2}}{(\sqrt{11} )^{2} -3^{2} } =\\\\\\=\frac{11-6\sqrt{11} +9}{11-9} =\frac{20-6\sqrt{11} }{2} =10-3\sqrt{11}$

Answer 2

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