Question

[Lê An Nguyễn]

1. Rút gọn biểu thức sau:

a, \(\sqrt{4+\sqrt{10+2\sqrt{5}}}+\sqrt{4-\sqrt{10+2\sqrt{5}}}\)

b, (\(\sqrt{3}-1\)) \(\sqrt{6+2\sqrt{2}\sqrt[]{3}-\sqrt{\sqrt{2}}+\sqrt{12}+\sqrt{18-128}}\)

Answer 1

[Phùng Khánh Linh]

\(a.A=\sqrt{4+\sqrt{10+2\sqrt{5}}}+\sqrt{4-\sqrt{10+2\sqrt{5}}}\)

⇔ \(A^2=\) \(\left(\sqrt{4+\sqrt{10+2\sqrt{5}}}+\sqrt{4-\sqrt{10+2\sqrt{5}}}\right)^2\)

⇔ \(A^2=4+\sqrt{10+2\sqrt{5}}+2\sqrt{\left(4+\sqrt{10+2\sqrt{5}}\right)\left(4-\sqrt{10+2\sqrt{5}}\right)}+4-\sqrt{10+2\sqrt{5}}\)⇔ \(A^2=8+2\sqrt{16-10-2\sqrt{5}}=8+2\sqrt{5-2\sqrt{5}+1}=8+2\sqrt{\left(\sqrt{5}-1\right)^2}\)

⇔ \(A^2=8+2\text{|}\sqrt{5}-1\text{|}\)

⇔ \(A^2=6+2\sqrt{5}=5+2\sqrt{5}+1=\left(\sqrt{5}+1\right)^2\)

⇔ \(\text{ |}A\text{ |}=\text{ |}\sqrt{5}+1\text{ |}\)

⇔ \(A=\sqrt{5}+1\)