Question

[Lê Kiều Trinh]

bài 5 Tính:

a) \(\sqrt{6-2\sqrt{5}}\)

b)\(\sqrt{7-4\sqrt{3}}\)

c)\(\sqrt{3-2\sqrt{2}}\)  -\(\sqrt{6-4\sqrt{2}}\)

d)\(\sqrt{13+30\sqrt{2+\sqrt{9+4\sqrt{2}}}}\)

Answer 1

[Akai Haruma]

Lời giải:

a. \(\sqrt{6-2\sqrt{5}}=\sqrt{5-2\sqrt{5}.\sqrt{1}+1}=\sqrt{(\sqrt{5}-1)^2}=\sqrt{5}-1\)

b. \(\sqrt{7-4\sqrt{3}}=\sqrt{4-2\sqrt{4}.\sqrt{3}+3}=\sqrt{(\sqrt{4}-\sqrt{3})^2}=\sqrt{4}-\sqrt{3}=2-\sqrt{3}\)

c.

\(\sqrt{3-2\sqrt{2}}-\sqrt{6-4\sqrt{2}}=\sqrt{2-2\sqrt{2}+1}-\sqrt{4-4\sqrt{2}+2}\)

\(=\sqrt{(\sqrt{2}-1)^2}-\sqrt{(\sqrt{4}-\sqrt{2})^2}\)

\(=|\sqrt{2}-1|-|\sqrt{4}-\sqrt{2}|=\sqrt{2}-1-(2-\sqrt{2})=2\sqrt{2}-3\)

d.

\(=\sqrt{13+30\sqrt{2+\sqrt{(\sqrt{8}+1)^2}}}=\sqrt{13+30\sqrt{2+\sqrt{8}+1}}\)

\(=\sqrt{13+30\sqrt{3+2\sqrt{2}}}=\sqrt{13+30\sqrt{(\sqrt{2}+1)^2}}\)

\(=\sqrt{13+30(\sqrt{2}+1)}=\sqrt{43+30\sqrt{2}}=\sqrt{18+2\sqrt{18.25}+25}\)

\(=\sqrt{(\sqrt{18}+\sqrt{25})^2}=\sqrt{18}+\sqrt{25}=5+3\sqrt{2}\)

Answer 2

[Nguyễn Lê Phước Thịnh]

a) \(\sqrt{6-2\sqrt{5}}=\sqrt{5}-1\)

b) \(\sqrt{7-4\sqrt{3}}=2-\sqrt{3}\)

c) \(\sqrt{3-2\sqrt{2}}-\sqrt{6-4\sqrt{2}}=\sqrt{2}-1-2+\sqrt{2}=-3+2\sqrt{2}\)

d) Ta có: \(\sqrt{13+30\sqrt{2+\sqrt{9+4\sqrt{2}}}}\)

\(=\sqrt{13+30\sqrt{2+1+2\sqrt{2}}}\)

\(=\sqrt{13+30\left(\sqrt{2}+1\right)}\)

\(=\sqrt{43+30\sqrt{2}}\)

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