Question

Giúp mình câu này với ạ,cảm ơn nhiều

Answer 1

\(\sqrt{2}\left(\sqrt{3+\sqrt{5}}\right)\left(\sqrt{5}-1\right)=\sqrt{6+2\sqrt{5}}\left(\sqrt{5}-1\right)\)

\(=\sqrt{\left(\sqrt{5}\right)^2+2.\sqrt{5}.1+1^2}\left(\sqrt{5}-1\right)=\sqrt{\left(\sqrt{5}+1\right)^2}\left(\sqrt{5}-1\right)\)

\(=\left|\sqrt{5}+1\right|\left(\sqrt{5}-1\right)=\left(\sqrt{5}-1\right)\left(\sqrt{5}+1\right)=4\)

Answer 2

1)Ta có: \(\sqrt{2}\cdot\sqrt{3+\sqrt{5}}\cdot\left(\sqrt{5}-1\right)\)

\(=\sqrt{6+2\sqrt{5}}\cdot\left(\sqrt{5}-1\right)\)

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

=5-1

=4