\(\sqrt{2-2.\frac{1}{2}\sqrt{2}+\frac{1}{4}}.\sqrt{8-2.2\sqrt{2}.\frac{1}{4}+\frac{1}{16}}=\sqrt{\left(\sqrt{2}-\frac{1}{2}\right)^2}\sqrt{\left(2\sqrt{2}-\frac{1}{4}\right)^2}\)

\(=\left(\sqrt{2}-\frac{1}{2}\right)\left(2\sqrt{2}-\frac{1}{4}\right)=\frac{33-10\sqrt{2}}{8}\)

\(\sqrt{2+2\sqrt{2}+1}.4\sqrt{\frac{288+2\sqrt{288}+1}{16}}=\sqrt{\left(\sqrt{2}+1\right)^2}.4\sqrt{\frac{\left(12\sqrt{2}+1\right)^2}{4^2}}\)

\(=\left(\sqrt{2}+1\right)\left(12\sqrt{2}+1\right)=25+13\sqrt{2}\)

\(\sqrt{28-10\sqrt{3}}=\sqrt{25-2.5\sqrt{3}+3}=\sqrt{\left(5-\sqrt{3}\right)^2}=5-\sqrt{3}\)

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

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

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

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

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

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

GIÚP MK NHANH NHA

a) \(\sqrt{\dfrac{59}{25}+\dfrac{6}{5}\sqrt{2}}=\sqrt{2+2.\dfrac{3}{5}\sqrt{2}+\dfrac{9}{25}}=\sqrt{\left(\sqrt{2}+\dfrac{3}{5}\right)^2}\)

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

b) \(\sqrt{\dfrac{129}{16}+\sqrt{2}}=\sqrt{8+2.2\sqrt{2}.\dfrac{1}{4}+\dfrac{1}{16}}\)

= \(\sqrt{\left(2\sqrt{2}+\dfrac{1}{4}\right)^2}\) = / \(2\sqrt{2}+\dfrac{1}{4}\) / = \(2\sqrt{2}+\dfrac{1}{4}\)

c) Tương tự , mình bận rồi , nếu chưa biết tẹo mk làm cho.

c) \(\sqrt{\dfrac{289+4\sqrt{72}}{16}}=\sqrt{\dfrac{289}{16}+\dfrac{1}{4}\sqrt{72}}=\sqrt{\dfrac{289}{16}+\dfrac{1}{4}.6\sqrt{2}}=\sqrt{18+2.\dfrac{1}{4}.3\sqrt{2}+\dfrac{1}{16}}=\sqrt{\left(3\sqrt{2}+\dfrac{1}{4}\right)^2}\) = / \(3\sqrt{2}+\dfrac{1}{4}\) / = \(3\sqrt{2}+\dfrac{1}{4}\)

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!