Question

(PUC )se x=$\frac{2}{3+2 \sqrt{2} } e y =$$\frac{56}{4- \sqrt{2} }$ , então x + y é igual ?

Answer 1

Oi   Karen :)
Veja aí se chegamos na mesma resposta. 

$x= \frac{2}{3+2 \sqrt{2} } . \frac{3-2 \sqrt{2}}{3-2 \sqrt{2}} = \frac{6-4 \sqrt{2} }{3^2-(2 \sqrt{2} )^2} = \frac{6-4 \sqrt{2} }{9-4*2}=\frac{6-4 \sqrt{2} }{1}= 6-4 \sqrt{2} \\ \\ y= \frac{56}{4- \sqrt{2} }. \frac{4+ \sqrt{2}}{4+ \sqrt{2}} = \frac{224+56 \sqrt{2} }{4^2-( \sqrt{2} )^2} = \frac{224+56 \sqrt{2} }{16-2}= \frac{224+56 \sqrt{2} }{14}=16+4 \sqrt{2} \\ \\ x+y=6-4 \sqrt{2}+16+4 \sqrt{2}=22$

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