Помогите с 192 номером, прошу умоляю очень надо, прошууу отзовитесь

решение во вложении.............

[latex] \frac{2+ \sqrt{6}}{2 \sqrt{2}+2 \sqrt{3}- \sqrt{6}-2}= \frac{2+ \sqrt{6}}{\sqrt{2}*2+ \sqrt{2}* \sqrt{2}* \sqrt{3}- \sqrt{6}-2}= \frac{2+ \sqrt{6}}{\sqrt{2}*(2+ \sqrt{2}*\sqrt{3})- (\sqrt{6}+2)}=[/latex] [latex]= \frac{2+ \sqrt{6}}{\sqrt{2}*(2+ \sqrt{2*3})-1*(\sqrt{6}+2)}= \frac{2+ \sqrt{6}}{\sqrt{2}*(2+ \sqrt{6})-1*(2+\sqrt{6})}= \frac{1*(2+ \sqrt{6})}{(\sqrt{2}-1)*(2+ \sqrt{6})}=[/latex] [latex]= \frac{1}{\sqrt{2}-1}=\frac{1*( \sqrt{2}+1)}{(\sqrt{2}-1)*( \sqrt{2}+1)}=\frac{1*( \sqrt{2}+1)}{(\sqrt{2})^2-1^2}=\frac{\sqrt{2}+1}{2-1}=1+\sqrt{2}[/latex] ------------------------- [latex] \sqrt{ \frac{2+ \sqrt{2} }{2- \sqrt{2} }* \sqrt{ \frac{2-\sqrt{2} }{2+\sqrt{2} }} } = \sqrt{( \sqrt{\frac{2+ \sqrt{2} }{2- \sqrt{2} }} )^2* \sqrt{ \frac{ 2-\sqrt{2} }{2+\sqrt{2} }} } =[/latex] [latex]=\sqrt{\sqrt{\frac{2+ \sqrt{2} }{2- \sqrt{2} }}*\sqrt{\frac{2+ \sqrt{2} }{2- \sqrt{2} }}* \sqrt{ \frac{ 2-\sqrt{2} }{2+\sqrt{2} }} } =\sqrt{\sqrt{\frac{2+ \sqrt{2} }{2- \sqrt{2} }*\frac{2+ \sqrt{2} }{2- \sqrt{2} }*\frac{ 2-\sqrt{2} }{2+\sqrt{2} }} } =[/latex] [latex]=\sqrt{\sqrt{\frac{2+ \sqrt{2} }{2- \sqrt{2}}}} =\sqrt{\sqrt{\frac{(2+ \sqrt{2})(2+ \sqrt{2})}{(2- \sqrt{2})(2+ \sqrt{2} )}}}} = \sqrt{\sqrt{\frac{(2+ \sqrt{2})^2}{2^2- (\sqrt{2})^2}}}} =[/latex] [latex]=\sqrt{\sqrt{\frac{(2+ \sqrt{2})^2}{4- 2}}}} = \sqrt{ \frac{2+ \sqrt{2} }{ \sqrt{2} } } = \sqrt{ \frac{ \sqrt{2} ( \sqrt{2}+1 )}{ \sqrt{2} } } = \sqrt{ \sqrt{2}+1 } [/latex]