a) \(\dfrac{\sqrt{2+\sqrt{3}}}{3+\sqrt{3}}=\dfrac{\sqrt{2}.\sqrt{2+\sqrt{3}}}{\sqrt{2}.\left(3+\sqrt{3}\right)}=\dfrac{\sqrt{4+2\sqrt{3}}}{\sqrt{6}.\left(\sqrt{3}+1\right)}=\dfrac{\sqrt{3}+1}{\sqrt{6}.\left(\sqrt{3}+1\right)}=\dfrac{1}{\sqrt{6}}.\)b) Đặt \(\sqrt{4-\sqrt{15}}+\sqrt{4+\sqrt{15}}-2\sqrt{3-\sqrt{5}}=A\)
Ta có:
\(\sqrt{2}A=\sqrt{8-\sqrt{15}}+\sqrt{8+\sqrt{15}}-2\sqrt{6-2\sqrt{5}}\)
\(\Rightarrow\sqrt{2}A=\left(\sqrt{\sqrt{5}-\sqrt{3}}\right)^2+\left(\sqrt{\sqrt{5}+\sqrt{3}}\right)^2-2\left(\sqrt{\sqrt{5}-1}\right)^2\)
\(\Rightarrow\sqrt{2}A=\sqrt{5}-\sqrt{3}+\sqrt{5}+\sqrt{3}-2\sqrt{5}-2\)
\(\Rightarrow\sqrt{2}A=-2\)
\(\Rightarrow A=-\sqrt{2}\)
a. \(\dfrac{\sqrt{2+\sqrt{3}}}{3+\sqrt{3}}=\dfrac{\sqrt{\left(\sqrt{\dfrac{3}{2}}+\sqrt{\dfrac{1}{2}}\right)^2}}{\sqrt{3}\left(\sqrt{3}+1\right)}=\dfrac{\dfrac{\sqrt{3}+1}{\sqrt{2}}}{\sqrt{3}\left(\sqrt{3}+1\right)}=\dfrac{\sqrt{3}+1}{\sqrt{2}}.\dfrac{1}{\sqrt{3}\left(\sqrt{3}+1\right)}=\dfrac{1}{\sqrt{6}}=\dfrac{\sqrt{6}}{6}\)
b. \(\sqrt{4-\sqrt{15}}+\sqrt{4+\sqrt{15}}-2\sqrt{3-\sqrt{5}}=\sqrt{\left(\sqrt{\dfrac{5}{2}}-\sqrt{\dfrac{3}{2}}\right)^2}+\sqrt{\left(\sqrt{\dfrac{5}{2}}+\sqrt{\dfrac{3}{2}}\right)^2}-2\sqrt{\left(\sqrt{\dfrac{5}{2}}-\sqrt{\dfrac{1}{2}}\right)^2}=\sqrt{\dfrac{5}{2}}-\sqrt{\dfrac{3}{2}}+\sqrt{\dfrac{5}{2}}+\sqrt{\dfrac{3}{2}}-2\sqrt{\dfrac{5}{2}}+2\sqrt{\dfrac{1}{2}}=\sqrt{2}\)
