Question

Să se determine dacă este numărul rațional valoarea expresiei numerice: $\frac{ \sqrt{2} }{ \sqrt{3+2 \sqrt{2} } } - \frac{ \sqrt{6-4 \sqrt{2} } }{2 \sqrt{2}-3 }$

Answer 1

[tex] \sqrt{3+ 2\sqrt{2} }= \sqrt{( \sqrt{2} )^2+2\cdot1\cdot \sqrt{2} +1^1}= \sqrt{( \sqrt{2}+1 )^2}=| \sqrt{2} +1|=\\ =\sqrt{2} +1[/tex]

[tex] \sqrt{6-4 \sqrt{2} } = \sqrt{( \sqrt{2})^2-2\cdot2\cdot \sqrt{2} +2^2}= \sqrt{(2- \sqrt{2} )^2}=|2-\sqrt{2}|=\\ =2-\sqrt{2} [/tex]

$2\sqrt{2}-3=-(3- 2\sqrt{2} )=-( (\sqrt{2})^2-2\cdot1\cdot\sqrt{2}+1^2 )=-(\sqrt{2}-1)^2$

[tex]n= \frac{\sqrt{2}}{\sqrt{2} +1} - \frac{2-\sqrt{2}}{-(\sqrt{2}-1)^2} = \frac{\sqrt{2}(\sqrt{2}-1)}{(\sqrt{2})^2-1^2}+ \frac{(2- \sqrt{2} )(\sqrt{2}+1)^2}{((\sqrt{2})^2-1^2)^2} =\\\\ = 2-\sqrt{2}+(2-\sqrt{2})(\sqrt{2}+1)^2=(2-\sqrt{2})(4+2\sqrt{2})=\\ =2(2-\sqrt{2})(2+\sqrt{2})=2(2^2-(\sqrt{2})^2)=4 \in Q[/tex]