Question

Вычислить:
$\sqrt{ (\frac{1+ \sqrt{2} }{1- \sqrt{2}})^{2}+2+(\frac{1- \sqrt{2} }{1+ \sqrt{2}})^{2} }$

Answer 1

$\sqrt{[(1+ \sqrt{2})/(1- \sqrt{2} )+(1- \sqrt{2}/(1+ \sqrt{2} )]^4 } =$$[(1+ \sqrt{2} )/(1- \sqrt{2} )+(1- \sqrt{2}) /(1+ \sqrt{2} )]^2=$$(1+ \sqrt{2} )^2/(1- \sqrt{2} )^2+2+(1- \sqrt{2} )^2/(1+ \sqrt{2} )^2$$=(3+2 \sqrt{2} )/(3-2 \sqrt{2} )+2+(3-2 \sqrt{2} )/(3+2 \sqrt{2} )=$$(9+12 \sqrt{2} +8+18-16+9-12 \sqrt{2} +8)/(9-8)=36/1=36$

Answer 2

Возведем в квадрат подкоренные значения
$\sqrt{ \frac{1+2 \sqrt{2}+2 }{1-2 \sqrt{2}+2 }+ \frac{1-2 \sqrt{2}+2 }{1+2 \sqrt{2}+2 } }$
$\sqrt{ \frac{3+2 \sqrt{2} }{3-2 \sqrt{2} }+ \frac{3-2 \sqrt{2} }{3+2 \sqrt{2} }+2 }$
Приводим к общему знаменателю
$\sqrt{ \frac{9+12 \sqrt{2}+8+9-12 \sqrt{2}+8 }{9-4*2} }$
$\sqrt{18+16+2}$=6