Question

(PUC-MG) Se x = $\frac{2}{3+2 \sqrt{2} }$ e y = $\frac{56}{4- \sqrt{2} }$, então x + y é igual a :
resposta: 22

Answer 1

$\frac{2}{3+2 \sqrt{2} } + \frac{56}{4- \sqrt{2} } \\ \\ \frac{2}{(3+2 \sqrt{2}) }. \frac{(3-2 \sqrt{2})}{(3-2 \sqrt{2}) }+ \frac{56}{(4- \sqrt{2}) }.\frac{(4+ \sqrt{2})}{(4+ \sqrt{2}) } \\ \\ \frac{6-4 \sqrt{2} }{3^2-(2 \sqrt{2} )^2} + \frac{224+56 \sqrt{2} }{4^2-( \sqrt{2} )^2} \\ \\ \frac{6-4 \sqrt{2} }{9-8} + \frac{224+56 \sqrt{2} }{16-2} \\ \\ \frac{6-4 \sqrt{2} }{1} + \frac{224+56 \sqrt{2} }{14} \\ \\ 6-4 \sqrt{2}+ \frac{14(16+4 \sqrt{2} )}{14} \\ \\ 6-4 \sqrt{2}+16+4 \sqrt{2}$
$6+16=22$