GIÚP MK NHANH NHA

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!

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

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

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

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

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

Học dỏi nha :)) 
~ Good luck ~

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

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

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

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

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

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

\(A=\sqrt{9-6\sqrt{7}+7}+\sqrt{3-2\sqrt{21}+7}\)

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

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

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

\(B=\sqrt{25+2\sqrt{75}+3}+\sqrt{16-2\sqrt{48}+3}\)

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

\(=5+\sqrt{3}+4-\sqrt{3}\)

\(=9\)

\(C=\sqrt{\sqrt{5}-\sqrt{3-\sqrt{20-12\sqrt{5}+9}}}\)

\(=\sqrt{\sqrt{5}-\sqrt{3-\sqrt{\left(2\sqrt{5}-3\right)^2}}}\)

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

\(=\sqrt{\sqrt{5}-\sqrt{5-2\sqrt{5}+1}}\)

\(=\sqrt{\sqrt{5}-\sqrt{\left(\sqrt{5}-1\right)^2}}\)

\(=\sqrt{\sqrt{5}-\sqrt{5}+1}\)
\(=\sqrt{1}\\ =1\)

b) \(\sqrt{12-3\sqrt{7}}-\sqrt{12+3\sqrt{7}}\)

= \(\sqrt{3.4-3\sqrt{7}}-\sqrt{3.4+3\sqrt{7}}\)

= \(\sqrt{3.\left(4-\sqrt{7}\right)}-\sqrt{3.\left(4+\sqrt{7}\right)}\)

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

= \(\sqrt{3}.\left(\sqrt{4-\sqrt{7}}-\sqrt{4+\sqrt{7}}\right)\)

\(\)≈ \(-2,449\)

\(\sqrt{\dfrac{13}{4}+\sqrt{3}}-\sqrt{\dfrac{7}{4}-\sqrt{3}}\)

= \(\sqrt{\dfrac{13}{4}+\dfrac{4\sqrt{3}}{4}}-\sqrt{\dfrac{7}{4}-\dfrac{4\sqrt{3}}{4}}\)

= \(\sqrt{\dfrac{13+4\sqrt{3}}{4}}-\sqrt{\dfrac{7-4\sqrt{3}}{4}}\)

= \(\dfrac{\sqrt{13+4\sqrt{3}}}{\sqrt{4}}-\dfrac{\sqrt{7-4\sqrt{3}}}{\sqrt{4}}\)

= \(\dfrac{\sqrt{13+4\sqrt{3}}-\sqrt{7-4\sqrt{3}}}{\sqrt{4}}\)

≈ \(2,098\)

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

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

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

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

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