Question

Tính

B=\(\sqrt{3-\sqrt{5}}-\sqrt{3+\sqrt{5}}\)

C=\(\left(\sqrt{6}+\sqrt{10}\right).\sqrt{4-\sqrt{15}}\)

D=\(\left(3+\sqrt{5}\right).\left(\sqrt{10}-\sqrt{2}\right).\sqrt{3-\sqrt{5}}\)

Answer 1

\(1.B=\sqrt{3-\sqrt{5}}-\sqrt{3+\sqrt{5}}=\dfrac{\sqrt{5-2\sqrt{5}+1}-\sqrt{5+2\sqrt{5}+1}}{\sqrt{2}}=\dfrac{\sqrt{5}-1-\sqrt{5}-1}{\sqrt{2}}=-\dfrac{2}{\sqrt{2}}=-\sqrt{2}\) \(2.C=\left(\sqrt{6}+\sqrt{10}\right)\sqrt{4-\sqrt{15}}=\sqrt{2}\left(\sqrt{3}+\sqrt{5}\right)\sqrt{4-\sqrt{15}}=\left(\sqrt{3}+\sqrt{5}\right)\sqrt{5-2\sqrt{5}.\sqrt{3}+3}=\left(\sqrt{3}+\sqrt{5}\right)\left(\sqrt{5}-\sqrt{3}\right)=5-3=2\) \(3.D=\left(3+\sqrt{5}\right)\left(\sqrt{10}-\sqrt{2}\right)\sqrt{3-\sqrt{5}}=\left(3+\sqrt{5}\right)\left(\sqrt{5}-1\right).\sqrt{5-2\sqrt{5}+1}=\left(3+\sqrt{5}\right)\left(\sqrt{5}-1\right)^2=2\left(3+\sqrt{5}\right)\left(3-\sqrt{5}\right)=2\left(9-5\right)=2.4=8\)