9) \(\sqrt{2-\sqrt{3}}\left(\sqrt{5}+\sqrt{2}\right)=\sqrt{\dfrac{4-2\sqrt{3}}{2}}\left(\sqrt{5}+\sqrt{2}\right)=\sqrt{\dfrac{\left(\sqrt{3}-1\right)^2}{2}}\left(\sqrt{5}+\sqrt{2}\right)=\dfrac{\left(\sqrt{3}-1\right)\left(\sqrt{5}+\sqrt{2}\right)}{\sqrt{2}}\)
10) \(\sqrt{3-\sqrt{5}}+\sqrt{3+\sqrt{5}}=\dfrac{\sqrt{6-2\sqrt{5}}+\sqrt{6+2\sqrt{5}}}{\sqrt{2}}=\dfrac{\sqrt{\left(\sqrt{5}-1\right)^2}+\sqrt{\left(\sqrt{5}+1\right)^2}}{\sqrt{2}}=\dfrac{2\sqrt{5}}{\sqrt{2}}=\sqrt{10}\)
11) Ta có:
\(A=\sqrt{4+\sqrt{10+2\sqrt{5}}}+\sqrt{4-\sqrt{10+2\sqrt{5}}}\)
\(A^2=4+\sqrt{10+2\sqrt{5}}+4-\sqrt{10+2\sqrt{5}}+2\sqrt{\left(4+\sqrt{10+2\sqrt{5}}\right)\left(4-\sqrt{10+2\sqrt{5}}\right)}\\ =8+2\sqrt{16-10-2\sqrt{5}}=8+2\sqrt{6-2\sqrt{5}}=8+2\sqrt{\left(\sqrt{5}-1\right)^2}=8+2\left(\sqrt{5}-1\right)=6+2\sqrt{5}=\left(\sqrt{5}+1\right)^2\)
\(A=\sqrt{5}+1\left(A>0\right)\)
