\(x=\frac{\left(\sqrt{3}-1\right)\sqrt[3]{10+6\sqrt{3}}}{\sqrt{21+4\sqrt{5}}+3}=\frac{\left(\sqrt{3}-1\right)\cdot\left(\sqrt{3}+1\right)}{2\sqrt{5}+1+3}=\frac{2}{2\sqrt{5}+4}=\sqrt{5}-2\)

\(A=\left(x^2-4x-2\right)^{2015}\)

Xong chắc thay vào nhưng để nguyên số mũ ko tính ra

a)\(\left(\sqrt{21}+7\right)\cdot\sqrt{10-2\sqrt{21}}\)

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

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

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

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

\(=\left(7+\sqrt{14}\right)\sqrt{\left(\sqrt{7}-\sqrt{2}\right)^2}\)

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

\(=\sqrt{7}\left(7-2\right)=5\sqrt{7}\)

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!

a) Ta có: \(A=\sqrt{8-2\sqrt{15}}\cdot\left(\sqrt{3}+\sqrt{5}\right)-\left(\sqrt{45}-\sqrt{20}\right)\)

\(=\sqrt{5-2\cdot\sqrt{5}\cdot\sqrt{3}+3}\cdot\left(\sqrt{5}+\sqrt{3}\right)-\sqrt{5}\left(\sqrt{9}-\sqrt{4}\right)\)

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

\(=\left|\sqrt{5}-\sqrt{3}\right|\cdot\left(\sqrt{5}+\sqrt{3}\right)-\sqrt{5}\)

\(=\left(\sqrt{5}-\sqrt{3}\right)\cdot\left(\sqrt{5}+\sqrt{3}\right)-\sqrt{5}\)(Vì \(\sqrt{5}>\sqrt{3}\))

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

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

b) Ta có: \(B=\left(\frac{\sqrt{21}-\sqrt{3}}{\sqrt{7}-1}-\frac{\sqrt{15}-\sqrt{3}}{1-\sqrt{5}}\right)\left(\frac{1}{2}\sqrt{6}-\sqrt{\frac{3}{2}}+3\sqrt{\frac{2}{3}}\right)\)

\(=\left(\frac{\sqrt{3}\left(\sqrt{7}-1\right)}{\sqrt{7}-1}+\frac{\sqrt{3}\left(\sqrt{5}-1\right)}{\sqrt{5}-1}\right)\left(\sqrt{\frac{3}{2}}-\sqrt{\frac{3}{2}}+\sqrt{6}\right)\)

\(=\sqrt{3}+\sqrt{3}+\sqrt{6}\)

\(=2\sqrt{3}+\sqrt{6}\)

c) Ta có: \(C=2\sqrt{3}+\sqrt{7-4\sqrt{3}}+\left(\sqrt{\frac{1}{3}}-\sqrt{\frac{4}{3}}+\sqrt{3}\right):\sqrt{3}\)

\(=2\sqrt{3}+\sqrt{4-2\cdot2\cdot\sqrt{3}+3}+\sqrt{\frac{1}{3}:3}-\sqrt{\frac{4}{3}:3}+\sqrt{3:3}\)

\(=2\sqrt{3}+\sqrt{\left(2-\sqrt{3}\right)^2}+\sqrt{\frac{1}{9}}-\sqrt{\frac{4}{9}}+\sqrt{1}\)

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

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

\(=\sqrt{3}+\frac{8}{3}\)

d) Ta có: \(D=\left(\frac{5+\sqrt{5}}{5-\sqrt{5}}+\frac{5-\sqrt{5}}{5+\sqrt{5}}\right):\frac{1}{\sqrt{7-4\sqrt{3}}}\)

\(=\left(\frac{\left(5+\sqrt{5}\right)^2+\left(5-\sqrt{5}\right)^2}{\left(5-\sqrt{5}\right)\left(5+\sqrt{5}\right)}\right)\cdot\sqrt{4-2\cdot2\cdot\sqrt{3}+3}\)

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

\(=\frac{60}{20}\cdot\left|2-\sqrt{3}\right|\)

\(=3\cdot\left(2-\sqrt{3}\right)\)(Vì \(2>\sqrt{3}\))

\(=6-3\sqrt{3}\)

Câu 1,2,3 Ez quá rồi :3

Câu 4:

Tổng quát:

\(\frac{1}{\sqrt{a}+\sqrt{a+1}}=\frac{\sqrt{a}-\sqrt{a+1}}{a-a-1}=\sqrt{a+1}-\sqrt{a}.\) Game là dễ :v

Câu 5 ko khác câu 4 lắm :v

Câu 5: 

Tổng quát:

\(\frac{1}{\sqrt{a}-\sqrt{a+1}}=\frac{\sqrt{a}+\sqrt{a+1}}{a-a-1}=-\sqrt{a}-\sqrt{a+1}.\) Game là dễ :v

Sao làm hổng ai bảo đú.n/g vậy :(((

a)=\(\sqrt{3-\sqrt{5}}\).\(\sqrt{3+\sqrt{5}}\).\(\sqrt{2}\)(\(\sqrt{5}\)-\(1\))\(\sqrt{3+\sqrt{5}}\)=2\(\sqrt{2}\) \(\sqrt{\left(\sqrt{5}-1\right)^2.\left(3+\sqrt{5}\right)}\)  =2\(\sqrt{2}\) .\(\sqrt{\left(6-2\sqrt{5}\right)\left(3+\sqrt{5}\right)}\) =2\(\sqrt{2}\)\(\sqrt{8}\)  =8

b)A²=8+2 căn[\(\left(4+\sqrt{10+2\sqrt{5}}\right)\left(4-\sqrt{10+2\sqrt{5}}\right)\)]=8+2\(\sqrt{6-2\sqrt{5}}\)=8+2(\(\sqrt{5}\)-1)=6+2\(\sqrt{5}\)=(\(\sqrt{5}+1\))² =>A=\(\sqrt{5}\)+1

c)C=\(\frac{2\sqrt{3}}{6}\)+\(\frac{\sqrt{2}}{6}\)-\(\frac{2\sqrt{\left(\sqrt{3}-\sqrt{2}\right)^2}}{6}\)=\(\frac{2\sqrt{3}+\sqrt{2}-2\left(\sqrt{3}-\sqrt{2}\right)}{6}\)=\(\frac{3\sqrt{2}}{6}\)=\(\frac{1}{\sqrt{2}}\)