Question

[Cô gái thất thường (Ánh Dương)]

Bài 1. thực hiện phép tính

a) \(\sqrt{\frac{5+\sqrt{21}}{5-\sqrt{21}}}+\sqrt{\frac{5-\sqrt{21}}{5+\sqrt{21}}}\)                                      b) \(\sqrt{\frac{4+\sqrt{7}}{4-\sqrt{7}}}+\sqrt{\frac{4-\sqrt{7}}{4+\sqrt{7}}}\)

Bài 2.  Tính:a) \(M=\left(2\sqrt{2}\right)\sqrt{2+4\sqrt{3+\sqrt{2}+\sqrt{11-6\sqrt{2}}}}\)

b) \(\frac{1}{\sqrt{25}+\sqrt{24}}+\frac{1}{\sqrt{24}+\sqrt{23}}+...+\frac{1}{\sqrt{2}+\sqrt{1}}=4\)

c)

\(\frac{\left(5\sqrt{3}+\sqrt{50}\right)\left(5-\sqrt{24}\right)}{\sqrt{75}0-5\sqrt{2}}\)

Answer 1

[ミ★kͥ-yͣeͫt★彡]

\(\frac{1}{\sqrt{25}+\sqrt{24}}+\frac{1}{\sqrt{24}+\sqrt{23}}+...+\frac{1}{\sqrt{2}+\sqrt{1}}=4\)

\(\Leftrightarrow\sqrt{25}-\sqrt{24}+\sqrt{24}-\sqrt{23}+...+\sqrt{2}-\sqrt{1}=4\)

\(\Leftrightarrow\sqrt{25}-\sqrt{1}=4\Leftrightarrow5-1=4\)(đúng)

Vậy \(\frac{1}{\sqrt{25}+\sqrt{24}}+\frac{1}{\sqrt{24}+\sqrt{23}}+...+\frac{1}{\sqrt{2}+\sqrt{1}}=4\)(đpcm)

Answer 2

[ミ★kͥ-yͣeͫt★彡]

\(M=\left(2\sqrt{2}\right)\sqrt{2+4\sqrt{3+\sqrt{2}+\sqrt{11-6\sqrt{2}}}}\)

\(=\left(2\sqrt{2}\right)\sqrt{2+4\sqrt{3+\sqrt{2}+\sqrt{2-6\sqrt{2}+9}}}\)

\(=\left(2\sqrt{2}\right)\sqrt{2+4\sqrt{3+\sqrt{2}+\sqrt{\left(3-\sqrt{2}\right)^2}}}\)

\(=\left(2\sqrt{2}\right)\sqrt{2+4\sqrt{3+\sqrt{2}+3-\sqrt{2}}}\)

\(=\left(2\sqrt{2}\right)\sqrt{2+4\sqrt{6}}\)

\(=\sqrt{16+32\sqrt{6}}\)