2-3-2-3-2-

Question Number 145224 by mathdanisur last updated on 03/Jul/21

\sqrt{2 \sqrt{3} + 2} - \sqrt{\sqrt{3} - \sqrt{2}} = ?

Answered by Olaf_Thorendsen last updated on 03/Jul/21

{(\sqrt{3} + \sqrt{2} + 1)}^{2} = 6 + 2 \sqrt{6} + 2 \sqrt{3} + 2 \sqrt{2}

x = \sqrt{2 \sqrt{3} + 2} - \sqrt{\sqrt{3} - \sqrt{2}}

x = \frac{1}{\sqrt{\sqrt{3} + \sqrt{2}}} (\sqrt{(2 \sqrt{3} + 2) (\sqrt{3} + \sqrt{2})} - 1)

x = \frac{1}{\sqrt{\sqrt{3} + \sqrt{2}}} (\sqrt{6 + 2 \sqrt{6} + 2 \sqrt{3} + 2 \sqrt{2}} - 1)

x = \frac{1}{\sqrt{\sqrt{3} + \sqrt{2}}} (\sqrt{{(\sqrt{3} + \sqrt{2} + 1)}^{2}} - 1)

x = \frac{1}{\sqrt{\sqrt{3} + \sqrt{2}}} (\sqrt{3} + \sqrt{2} + 1 - 1)

x = \frac{1}{\sqrt{\sqrt{3} + \sqrt{2}}} (\sqrt{3} + \sqrt{2})

x = \sqrt{\sqrt{3} + \sqrt{2}}