Question

Tính

a, \(A=\left(\sqrt{6}+\sqrt{10}\right)\sqrt{4-\sqrt{15}}\)

b, \(B=\left(3+\sqrt{5}\right)\left(\sqrt{10}-\sqrt{2}\right)\sqrt{3-\sqrt{5}}\)

Answer 1

a) \(A=\left(\sqrt{6}+\sqrt{10}\right)\sqrt{4-\sqrt{15}}=\left(\sqrt{3}+\sqrt{5}\right)\sqrt{2}\cdot\sqrt{4-\sqrt{15}}=\left(\sqrt{3}+\sqrt{5}\right)\cdot\left(\sqrt{5}-\sqrt{3}\right)=2\)

b) \(B=\left(3+\sqrt{5}\right)\left(\sqrt{10}-\sqrt{2}\right)\sqrt{3-\sqrt{5}}=\left(3+\sqrt{5}\right)\left(\sqrt{5}-1\right)\sqrt{2}\cdot\sqrt{3-\sqrt{5}}=\left(3+\sqrt{5}\right)\left(\sqrt{5}-1\right)\left(\sqrt{5}-1\right)=\dfrac{\left(\sqrt{5}+1\right)^2\cdot\left(\sqrt{5}-1\right)^2}{2}=\dfrac{4^2}{2}=8\)

Answer 2

Sao lại bằng \(\left(\sqrt{3}+\sqrt{5}\right)\left(\sqrt{5}-\sqrt{3}\right)\)