Question

ПОМОГИТЕЕЕ ПОЖАЛУЙСТА РЕШИТЬ

Answer 1

$\frac{1}{\sqrt{2-\sqrt3}+1}-\frac{1}{\sqrt{2+\sqrt3}-1} =\frac{1}{\sqrt{(\sqrt{\frac{3}{2}}-\sqrt{\frac{1}{2}})^2}+1}-\frac{1}{\sqrt{(\sqrt{\frac{3}{2}}+\sqrt{\frac{1}{2}})^2}-1}=\\\\=\frac{1}{\sqrt{\frac{3}{2}}-\sqrt{\frac{1}{2}}+1}-\frac{1}{\sqrt{\frac{3}{2}}+\sqrt{\frac{1}{2}}+1}=\frac{\sqrt2}{\sqrt3-1+\sqrt2}-\frac{\sqrt2}{\sqrt3+1+\sqrt2}=\\\\=\frac{\sqrt2(\sqrt3+\sqrt2+1)-\sqrt2(\sqrt3+\sqrt2-1)}{((\sqrt3+\sqrt2)-1)\cdot ((\sqrt3+\sqrt2)+1)}=\\\\=\frac{\sqrt6+2+\sqrt2-\sqrt6-2+\sqrt2}{(\sqrt3+\sqrt2)^2-1^2}$

$=\frac{2\sqrt2}{3+2\sqrt6+2-1}=\frac{2\sqrt{2}}{4+2\sqrt6}=\frac{\sqrt2}{2+\sqrt6}=\frac{\sqrt2}{\sqrt2(\sqrt2+\sqrt3)}=\\\\=\frac{1}{\sqrt3+\sqrt2}=\frac{\sqrt3-\sqrt2}{3-2}=\sqrt3-\sqrt2$

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