Помогите плз..........

$\frac{1}{ \sqrt{3} +( \sqrt{2}+1 )} = \frac{1*[\sqrt{3} -( \sqrt{2}+1)]}{[ \sqrt{3} +( \sqrt{2}+1 )]*[\sqrt{3} -( \sqrt{2}+1)]}= \frac{\sqrt{3} -( \sqrt{2}+1)}{3-( \sqrt{2 } +1)^2}= \\ \\=\frac{\sqrt{3} -( \sqrt{2}+1)}{3-( 2+2 \sqrt{2}+1) }=\frac{\sqrt{3} -( \sqrt{2}+1)}{3-( 3+2 \sqrt{2}) }=\frac{\sqrt{3} -( \sqrt{2}+1)}{3-3-2 \sqrt{2} }=-\frac{\sqrt{3} -( \sqrt{2}+1)}{2 \sqrt{2} }= \\ \\ = \frac{ \sqrt{2} +1- \sqrt{3} }{2 \sqrt{2} } =$
$\frac{ (\sqrt{2} +1- \sqrt{3})* \sqrt{2} }{2 \sqrt{2} * \sqrt{2} } = \frac{2 + \sqrt{2} - \sqrt{6} }{2*2 } = \frac{2 + \sqrt{2} - \sqrt{6} }{4 } \\ \\ OTBET: C$