Question

Passo a passo para o resultado a baixo :

Answer 1

$\frac{ (\frac{1}{ \sqrt{2} } +1)^2-1}{(\frac{1}{ \sqrt{2} } -1)^2+1} \ \ \ \ \ \ \ \ \ \ (a+b)^2=a^2+2ab+b^2 \\ \\ \frac{( (\frac{1}{ \sqrt{2} } )^2+2. \frac{1}{ \sqrt{2} }.1+1^2)-1 }{( (\frac{1}{ \sqrt{2} } )^2-2. \frac{1}{ \sqrt{2} }.1+1^2)+1} = \frac{\frac{1}{ 2 }+ \frac{2}{ \sqrt{2} }+1-1 }{\frac{1}{ 2 }- \frac{2}{ \sqrt{2} }+1+1} = \frac{ \frac{1}{2}+ \frac{2}{ \sqrt{2} } }{ \frac{5}{2}- \frac{2}{ \sqrt{2} } }= \frac{ \frac{ \sqrt{2}+4 }{ 2\sqrt{2} } }{\frac{5 \sqrt{2}-4 }{ 2\sqrt{2} } }$
$\frac{ \sqrt{2}+4 }{ 2\sqrt{2} }.\frac{2 \sqrt{2} }{5 \sqrt{2}-4 }= \frac{\sqrt{2}+4}{5 \sqrt{2}-4 } \ \ \ \ racionalizando\\ \\ \frac{\sqrt{2}+4}{5 \sqrt{2}-4 } . \frac{5 \sqrt{2}+4 }{5 \sqrt{2}+4 } = \frac{5 (\sqrt{2})^2+4 \sqrt{2}+20 \sqrt{2}+4^2 }{(5 \sqrt{2} )^2-4^2} = \frac{5*2+16+24 \sqrt{2} }{25*2-16} = \frac{26+24 \sqrt{2} }{34}= \\ \\ \frac{2(13+12 \sqrt{2} )}{2(17)}= \frac{13+12 \sqrt{2} }{17}$