\(\sqrt{\frac{9-4\sqrt{2}}{4}}=\frac{\sqrt{\left(2\sqrt{2}-1\right)^2}}{2}=\frac{2\sqrt{2}-1}{2}\)

\(\sqrt{\frac{129+16\sqrt{2}}{16}}=\sqrt{\frac{\left(8\sqrt{2}+1\right)^2}{16}}=\frac{8\sqrt{2}+1}{4}\)

\(\sqrt{3+2\sqrt{2}}=\sqrt{\left(\sqrt{2}+1\right)^2}=\sqrt{2}+1\)

\(\sqrt{\frac{289+4\sqrt{72}}{16}}=\frac{\sqrt{\left(12\sqrt{2}+1\right)^2}}{4}=\frac{12\sqrt{2}+1}{4}\)

\(\sqrt{8+2\sqrt{15}}=\sqrt{\left(\sqrt{5}+\sqrt{3}\right)^2}=\sqrt{5}+\sqrt{3}\)

Học dỏi nha :)) 
~ Good luck ~

\(\sqrt{\frac{289+4\sqrt{72}}{16}}+\sqrt{\frac{129}{16}+\sqrt{2}}\)

\(=\sqrt{\frac{288+2\times12\sqrt{2}+1}{4^2}}+\sqrt{\frac{128+2\sqrt{12}+1}{4^2}}\)

\(=\sqrt{\frac{\left(\sqrt{288}+1\right)^2}{4^2}}+\sqrt{\frac{\left(\sqrt{128}+1\right)^2}{4^2}}\)

\(=\frac{\sqrt{288}+1}{4}+\frac{\sqrt{128}+1}{4}\)

\(=\frac{12\sqrt{2}+8\sqrt{2}+2}{4}\)

\(=\frac{1+10\sqrt{2}}{2}\)

GIÚP MK NHANH NHA

\(\sqrt{16-6\sqrt{7}}=\sqrt{9-2.3\sqrt{7}+7}=\sqrt{\left(3-\sqrt{7}\right)^2}=3-\sqrt{7};\sqrt{10-2\sqrt{21}}=\sqrt{3-2\sqrt{3}\sqrt{7}+7}=\sqrt{\left(\sqrt{7}-\sqrt{3}\right)^2}=\sqrt{7}-\sqrt{3}\Rightarrow\sqrt{16-6\sqrt{7}}+\sqrt{10-2\sqrt{21}}=3-\sqrt{3}\)

1. Đặt A =\(\sqrt{\frac{129}{16}+\sqrt{2}}\)

\(\sqrt{16}\)A = \(\sqrt{129+16\sqrt{2}}\)

4A = \(\sqrt{\left(8\sqrt{2}+1\right)^2}\)

4A = \(8\sqrt{2}+1\)

⇒ A = \(\frac{\text{​​}8\sqrt{2}+1}{4}\)= \(2\sqrt{2}\) + \(\frac{1}{4}\)

2. Đặt B = \(\sqrt{\frac{289+4\sqrt{72}}{16}}\)

\(\sqrt{16}\)B = \(\sqrt{289+24\sqrt{2}}\)

4B = \(\sqrt{\left(12\sqrt{2}+1\right)^2}\)

4B = \(12\sqrt{2}+1\)

⇒ B = \(\frac{12\sqrt{2}+1}{4}\)= \(3\sqrt{2}+\frac{1}{4}\)

3. \(\sqrt{2-\sqrt{3}}\). \(\left(\sqrt{6}+\sqrt{2}\right)\)

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

= \(\sqrt{4-2\sqrt{3}}\) . \(\left(\sqrt{3}+1\right)\)

= \(\sqrt{\left(\sqrt{3}-1\right)^2}\) . \(\left(\sqrt{3}+1\right)\)

= \(\left(\sqrt{3}-1\right)\). \(\left(\sqrt{3}+1\right)\)

= \(\left(\sqrt{3}\right)^2\) - 1²

= 3 - 1

= 2

4. \(\left(\sqrt{21}+7\right)\). \(\sqrt{10-2\sqrt{21}}\)

= \(\left(\sqrt{21}+7\right)\) . \(\sqrt{\left(\sqrt{7}-\sqrt{3}\right)^2}\)

= \(\sqrt{7}\left(\sqrt{3}+\sqrt{7}\right)\) . \(\left(\sqrt{7}-\sqrt{3}\right)\)

= \(\sqrt{7}\) \(\left[\left(\sqrt{7}\right)^2-\left(\sqrt{3}\right)^2\right]\)

= \(\sqrt{7}\) . (7 - 3)

= 4\(\sqrt{7}\)

5. \(2.\left(\sqrt{10}-\sqrt{2}\right)\). \(\sqrt{4+\sqrt{6-2\sqrt{5}}}\)

= \(2.\left(\sqrt{10}-\sqrt{2}\right)\) . \(\sqrt{4+\sqrt{5}-1}\)

= \(2.\left(\sqrt{10}-\sqrt{2}\right)\) . \(\sqrt{3+\sqrt{5}}\)

= \(\left(\sqrt{10}-\sqrt{2}\right)\) . \(\sqrt{12+4\sqrt{5}}\)

= \(\left(\sqrt{10}-\sqrt{2}\right)\) . \(\left(\sqrt{10}+\sqrt{2}\right)\)

= \(\left(\sqrt{10}\right)^2-\left(\sqrt{2}\right)^2\)

= 10 - 2

= 8

6. \(\left(4\sqrt{2}+\sqrt{30}\right)\). \(\left(\sqrt{5}-\sqrt{3}\right)\). \(\sqrt{4-\sqrt{15}}\)

= \(\sqrt{2}\)\(\left(4+\sqrt{15}\right)\) . \(\left(\sqrt{5}-\sqrt{3}\right)\) . \(\sqrt{4-\sqrt{15}}\)

= \(\left(4+\sqrt{15}\right)\) . \(\left(\sqrt{5}-\sqrt{3}\right)\) . \(\sqrt{8-2\sqrt{15}}\)

= \(\left(4+\sqrt{15}\right)\) . \(\left(\sqrt{5}-\sqrt{3}\right)\) . \(\left(\sqrt{5}-\sqrt{3}\right)\)

= \(\left(4+\sqrt{15}\right)\) . \(\left(\sqrt{5}-\sqrt{3}\right)^2\)

= \(\left(4+\sqrt{15}\right)\). \(\left(8-2\sqrt{15}\right)\)

= 32 - \(8\sqrt{15}\) + \(8\sqrt{15}\) - 30

= 2

7. \(\left(7-\sqrt{14}\right)\) . \(\sqrt{9-2\sqrt{14}}\)

= \(\sqrt{7}\) \(\left(\sqrt{7}-\sqrt{2}\right)\). \(\left(\sqrt{7}-\sqrt{2}\right)\)

= \(\sqrt{7}\). \(\left(\sqrt{7}-\sqrt{2}\right)^2\)

= \(\sqrt{7}\) . \(\left(9-2\sqrt{14}\right)\)

= 9\(\sqrt{7}\) - 14\(\sqrt{2}\)

TICK MÌNH NHA!

Bạn thông minh ghê!

\(b,\frac{2+\sqrt{3}}{1-\sqrt{4-2\sqrt{3}}}+\frac{2-\sqrt{3}}{1+\sqrt{4+2\sqrt{3}}}\)

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

\(=\frac{2+\sqrt{3}}{1-\sqrt{\left(\sqrt{3}-1\right)^2}}+\frac{2-\sqrt{3}}{1+\sqrt{\left(\sqrt{3}+1\right)^2}}\)

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

\(=\frac{2+\sqrt{3}}{2-\sqrt{3}}+\frac{2-\sqrt{3}}{2+\sqrt{3}}\)

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

\(=\frac{4+4\sqrt{3}+3+4-4\sqrt{3}+3}{4-3}\)

\(=14\)

\(a,\frac{\sqrt{2}+\sqrt{3}+\sqrt{6}+\sqrt{8}+\sqrt{16}}{\sqrt{2}+\sqrt{3}+\sqrt{4}}\)

\(=\frac{\sqrt{2}+\sqrt{3}+4+\sqrt{6}+\sqrt{8}}{\sqrt{2}+\sqrt{3}+2}\)

\(=\frac{\sqrt{2}+\sqrt{3}+2}{\sqrt{2}+\sqrt{3}+2}+\frac{\sqrt{2}.\sqrt{2}+\sqrt{2}.\sqrt{3}+\sqrt{2}.2}{\sqrt{2}+\sqrt{3}+2}\)

\(=1+\frac{\sqrt{2}\left(\sqrt{2}+\sqrt{3}+2\right)}{\sqrt{2}+\sqrt{3}+2}\)

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

a) \(\frac{\sqrt{6}+\sqrt{14}}{2\sqrt{3}+\sqrt{28}}=\frac{\sqrt{2}\left(\sqrt{3}+\sqrt{7}\right)}{2\left(\sqrt{3}+\sqrt{7}\right)}=\frac{1}{\sqrt{2}}\)

b) \(\frac{\sqrt{2}+\sqrt{3}+\sqrt{6}+\sqrt{8}+\sqrt{16}}{\sqrt{2}+\sqrt{3}+\sqrt{4}}\)

\(=\frac{\left(\sqrt{2}+\sqrt{3}+\sqrt{4}\right)+\left(\sqrt{4}+\sqrt{6}+\sqrt{8}\right)}{\sqrt{2}+\sqrt{3}+\sqrt{4}}\)

\(=\frac{\left(\sqrt{2}+\sqrt{3}+\sqrt{4}\right)\left(1+\sqrt{2}\right)}{\sqrt{2}+\sqrt{3}+\sqrt{4}}\)

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