9: Ta có: \(\left(1+\sqrt{2}-\sqrt{3}\right)\left(1+\sqrt{2}+\sqrt{3}\right)\)
\(=\left(\sqrt{2}+1\right)^2-3\)
\(=3+2\sqrt{2}-3=2\sqrt{2}\)
10: Ta có: \(\dfrac{\sqrt{15}-\sqrt{5}}{\sqrt{6}-\sqrt{2}}+2\sqrt{10}\)
\(=\dfrac{\sqrt{5}\left(\sqrt{3}-1\right)}{\sqrt{2}\left(\sqrt{3}-1\right)}+2\sqrt{10}\)
\(=\dfrac{\sqrt{10}}{2}+\dfrac{4\sqrt{10}}{2}=\dfrac{5\sqrt{10}}{2}\)
13)\(\dfrac{2\sqrt{10}+\sqrt{30}-2\sqrt{2}-\sqrt{6}}{2\sqrt{10}-2\sqrt{2}}=\dfrac{\sqrt{10}\left(2+\sqrt{3}\right)-\sqrt{2}\left(2+\sqrt{3}\right)}{2\left(\sqrt{10}-\sqrt{2}\right)}\)\(=\dfrac{\left(\sqrt{10}-\sqrt{2}\right)\left(2+\sqrt{3}\right)}{2\left(\sqrt{10}-\sqrt{2}\right)}=\dfrac{2+\sqrt{3}}{2}\)
14)sai đề? phải là \(\sqrt{3-\sqrt{5}}\)
\(=\dfrac{\sqrt{3-\sqrt{5}}\left(3+\sqrt{5}\right)}{2\sqrt{10}-2\sqrt{2}}=\dfrac{\sqrt{6-2\sqrt{5}}\left(3+\sqrt{5}\right)}{\sqrt{2}\left(2\sqrt{10}-2\sqrt{2}\right)}\)
\(=\dfrac{\sqrt{\left(\sqrt{5}-1\right)^2}\left(3+\sqrt{5}\right)}{4\left(\sqrt{5}-1\right)}=\dfrac{\left|\sqrt{5}-1\right|\left(3+\sqrt{5}\right)}{4\left(\sqrt{5}-1\right)}\)
\(=\dfrac{3+\sqrt{5}}{4}\)
15)\(\sqrt{\left(1-\sqrt{2016}\right)^2}.\sqrt{2017+2\sqrt{2016}}=\left|1-\sqrt{2016}\right|\sqrt{1+2\sqrt{2016}+2016}\)
\(=\left(\sqrt{2016}-1\right)\sqrt{\left(1+\sqrt{2016}\right)^2}=\left(\sqrt{2016}-1\right)\left(1+\sqrt{2016}\right)\)
\(=2015\)