Question

[Nhi Lê Nguyễn Bảo]

a/ \(\dfrac{\sqrt{3-\sqrt{5}}\left(3+\sqrt{5}\right)}{\sqrt{10}+\sqrt{2}}\)

b/ \(\sqrt{8\sqrt{3}}-2\sqrt{25\sqrt{12}}+4\sqrt{\sqrt{192}}\)

c/ \(\sqrt{2-\sqrt{3}}\left(\sqrt{5}+\sqrt{2}\right)\)

d/ \(\sqrt{3-\sqrt{5}}+\sqrt{3+\sqrt{5}}\)

e/ \(\dfrac{\left(\sqrt{5}+2\right)^2-8\sqrt{5}}{2\sqrt{5}-4}\)

Làm ơn, giúp mik với. Mik đang cần gấp lắm!

Answer 1

[Nguyễn Lê Phước Thịnh]

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

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

b: \(=\sqrt{\sqrt{3}}\left(2\sqrt{2}-2\cdot5\sqrt{2}+4\cdot8\sqrt{2}\right)\)

\(=\sqrt{\sqrt{3}}\cdot24\sqrt{2}\)

d: \(=\dfrac{\sqrt{6-2\sqrt{5}}+\sqrt{6+2\sqrt{5}}}{\sqrt{2}}\)

\(=\dfrac{\sqrt{5}-1+\sqrt{5}+1}{\sqrt{2}}=\dfrac{2\sqrt{5}}{\sqrt{2}}=\sqrt{10}\)