Question

Tính

\(Q=\left(3-\sqrt{5}\right)\sqrt{3+\sqrt{5}}+\left(3+\sqrt{5}\right)\sqrt{3-\sqrt{5}}\)

Answer 1

\(Q=\dfrac{\left(3-\sqrt{5}\right)\cdot\sqrt{6+2\sqrt{5}}+\left(3+\sqrt{5}\right)\sqrt{6-2\sqrt{5}}}{\sqrt{2}}\)

\(=\dfrac{\left(3-\sqrt{5}\right)\left(\sqrt{5}+1\right)+\left(3+\sqrt{5}\right)\left(\sqrt{5}-1\right)}{\sqrt{2}}\)

\(=\dfrac{3\sqrt{5}+3-5-\sqrt{5}+3\sqrt{5}-3+5-\sqrt{5}}{\sqrt{2}}\)

\(=\dfrac{4\sqrt{5}}{\sqrt{2}}=2\sqrt{10}\)

Answer 2

giúp mình với ạ

Answer 3

Answer 4