\(8,I=\sqrt{2}-\dfrac{\left(2-\sqrt{2}\right)\left(\sqrt{2}+1\right)}{2-1}=\sqrt{2}-\sqrt{2}=0\\ 9,J=1^2-\left(\dfrac{2+\sqrt{2}}{1+\sqrt{2}}\right)^2=1-\left(\dfrac{\left(2+\sqrt{2}\right)\left(\sqrt{2}-1\right)}{2}\right)^2\\ =1-\left(\sqrt{2}\right)^2=-1\\ 10,K=\dfrac{2\left(2+\sqrt{5}\right)-2\left(2-\sqrt{5}\right)}{\left(2-\sqrt{5}\right)\left(2+\sqrt{5}\right)}=\dfrac{4+2\sqrt{5}-4+2\sqrt{5}}{-1}=-4\sqrt{5}\)
\(11,L=\left(\dfrac{\sqrt{2}\left(\sqrt{3}-1\right)}{-\left(\sqrt{3}-1\right)}-\sqrt{3}\right):\dfrac{1}{\sqrt{2}-\sqrt{3}}\\ =-\left(\sqrt{2}+\sqrt{3}\right)\left(\sqrt{2}-\sqrt{3}\right)=-\left(-1\right)=1\\ 12,M=\dfrac{\sqrt{6}\left(\sqrt{3}-\sqrt{2}\right)}{\sqrt{3}-\sqrt{2}}\cdot6=6\sqrt{6}\\ 13,N=\dfrac{6\sqrt{5}+1+\sqrt{5}}{\sqrt{5}\left(1+\sqrt{5}\right)}=\dfrac{7\sqrt{5}+1}{5+\sqrt{5}}=\dfrac{\left(7\sqrt{5}+1\right)\left(5-\sqrt{5}\right)}{20}\\ =\dfrac{34\sqrt{5}-30}{20}=\dfrac{17\sqrt{5}-15}{10}\)
\(14,P=\dfrac{2+2\sqrt{2}-2+2\sqrt{2}}{\left(1-\sqrt{2}\right)\left(1+\sqrt{2}\right)}=\dfrac{4\sqrt{2}}{-1}=-4\sqrt{2}\\ 15,Q=\left(\dfrac{\sqrt{2}\left(\sqrt{3}-1\right)}{-\left(\sqrt{3}-1\right)}-\sqrt{5}\right)\left(\sqrt{5}-\sqrt{2}\right)\\ =-\left(\sqrt{2}+\sqrt{5}\right)\left(\sqrt{5}-\sqrt{2}\right)=-3\\ 16,R=\dfrac{14-8\sqrt{3}+14+8\sqrt{3}}{\left(7+4\sqrt{3}\right)\left(7-4\sqrt{3}\right)}=\dfrac{28}{1}=28\)
