Question

Sa se determine daca este un numar rational expressiei numerice:
$\frac{ \sqrt{2}+1 }{ \sqrt{3+2 \sqrt{2} } }$


$\frac{ \sqrt{2} }{ \sqrt{3+2 \sqrt{2} } }- \frac{ \sqrt{6-4 \sqrt{2} } }{2 \sqrt{2}-3 }$

Answer 1

$\frac{\sqrt{2}+1}{\sqrt{3+2\sqrt{2}}}= \frac{\sqrt{2}+1}{\sqrt{3+\sqrt{8}}} \\ \\ \\\sqrt{3+\sqrt{8}} \\ A=3 \\ B=8 \\ C^2=9-8=>C=1 \\ \\ \sqrt{3+\sqrt{8}}= \sqrt{\frac{3+1}{2}+\sqrt{\frac{3-1}{2}}}= \sqrt{\frac{4}{2}+ \sqrt{\frac{2}{2}}}= \sqrt{2}+1 \\ \\ \frac{\sqrt{2}+1}{\sqrt{3+\sqrt{8}}}= \frac{\sqrt{2}+1}{\sqrt{2}+1}=1$

$\frac{\sqrt{2}}{\sqrt3+2\sqrt{2}}= \frac{\sqrt{2}}{\sqrt{2}+1} \\ \\ \sqrt{6-4\sqrt{2}}= \sqrt{6-\sqrt{32}} \\ A=6 \\ B=32 \\ C^2=36-32 => C=2 \\ \\ \sqrt{\frac{6+2}{2}-\sqrt{\frac{6-2}{2}}}= \sqrt{ \frac{8}{2}-\sqrt{\frac{4}{2}}}= \sqrt{4}-\sqrt{2}= 2-\sqrt{2} \\ \\ \\ \frac{\sqrt{2}}{\sqrt{2}+1}}^{(\sqrt{2}-1}- \frac{2-\sqrt{2}}{2\sqrt{2}-3}}^{(2\sqrt{2}+3}= \frac{\sqrt{2}(\sqrt{2}-1)}{2-1} - \frac{(2-\sqrt{2})(2\sqrt{2}+3)}{8-9} \\ \\ \\ 2-\sqrt{2}+ (4\sqrt{2}+6-4-3\sqrt{2})= 2-\sqrt{2}+\sqrt{2}+2=2+2=4$