Question

$\sqrt{3 + \sqrt{5} } - \sqrt{3 - \sqrt{5} }$
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Answer 1

$\sqrt{3+\sqrt{5}}-\sqrt{3-\sqrt{5}} =A$

$( \sqrt{3+\sqrt{5}}-\sqrt{3 - \sqrt{5} })^2=A^2$

$(\sqrt{3+\sqrt{5}})^2+(\sqrt{3 -\sqrt{5}})^2-2*\sqrt{(3+\sqrt{5})(3-\sqrt{5})} =A^2$

$3+\sqrt{5}}+3 -\sqrt{5}-2*\sqrt{3^2-(\sqrt{5})^2} =A^2$

$6-2*\sqrt{9-5} =A^2$

$6-2*2 =A^2$

$A^2=2$

$A=б\sqrt{2}$

так как $\sqrt{3+\sqrt{5}} >\sqrt{3-\sqrt{5}} ,$  то

$A=\sqrt{2}$