Question

4 / (√2 + √(6 + 4 × √(2))
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Answer 1

$\frac{4}{ \sqrt{2} + \sqrt{6 + 4 \sqrt{2} } } = \frac{4}{ \sqrt{2} + \sqrt{4 + 4 \sqrt{2} + 2} } = \\ = \frac{4}{ \sqrt{2} + \sqrt{(2 + \sqrt{2}) ^{2} } } = \frac{4}{ \sqrt{2} + 2 + \sqrt{2} } = \frac{4}{2 \sqrt{2} + 2 } = \\ = \frac{4}{2( \sqrt{2} + 1)} = \frac{2}{ \sqrt{2} + 1} = \frac{2( \sqrt{2} - 1)}{( \sqrt{2} + 1)( \sqrt{2} - 1)} = \frac{2( \sqrt{2} - 1)}{2 - 1} = 2( \sqrt{2} - 1) = \\ = 2 \sqrt{2 } - 2$

Answer 2

$\frac{4}{\sqrt{2}+\sqrt{6+4\sqrt{2}}}=\frac{4}{\sqrt{2}+\sqrt{4+4\sqrt{2}+2}}=\frac{4}{\sqrt{2}+\sqrt{2^2+2*2\sqrt{2}+(\sqrt{2})^2}}= \\ \\ =\frac{4}{\sqrt{2}+\sqrt{(2+\sqrt{2})^2}}=\frac{4}{\sqrt{2}+|2+\sqrt{2}|}}=\frac{4}{\sqrt{2}+2+\sqrt{2}}}=\frac{4}{2(\sqrt{2}+1)}=\\ \\ =\frac{2(\sqrt{2}-1)}{(\sqrt{2}+1)(\sqrt{2}-1)}=\frac{2(\sqrt{2}-1)}{2-1}=2(\sqrt{2}-1)$