Question

(2+корень3)/(2+корень(4+2корень3))+(2-корень3)/(2-корень(4-2корень3))

Answer 1

$\frac{2+ \sqrt{3} }{2+ \sqrt{4+2 \sqrt{3} } } + \frac{2- \sqrt{3} }{2+ \sqrt{4-2 \sqrt{3} } }= \frac{2+ \sqrt{3} }{2+ \sqrt{1+3+2 \sqrt{3} } } + \frac{2- \sqrt{3} }{2+ \sqrt{1+3-2 \sqrt{3} } }=$

$=\frac{2+ \sqrt{3} }{2+ \sqrt{1^2+2*1* \sqrt{3} +( \sqrt{3} )^2} } + \frac{2- \sqrt{3} }{2+ \sqrt{1^2-2*1* \sqrt{3} +( \sqrt{3} )^2} }=$

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

$=\frac{(2+ \sqrt{3})(3-\sqrt{3})+ (2- \sqrt{3})(3+\sqrt{3})}{(3+ \sqrt{3})(3- \sqrt{3})} } = \frac{6-2 \sqrt{3} +3 \sqrt{3} -3+6+2 \sqrt{3}-3 \sqrt{3} -3 }{9-3} = \frac{6}{6} =1$