a.

\(\sqrt[3]{125}.\sqrt[3]{\frac{16}{10}}.\sqrt[3]{-0,5}=\sqrt[3]{125.\frac{16}{10}.(-0,5)}=\sqrt[3]{-100}\)

b.

\(=1+\frac{1}{\sqrt[3]{4}+\sqrt[3]{2}+1}=1+\frac{\sqrt[3]{2}-1}{(\sqrt[3]{2}-1)(\sqrt[3]{4}+\sqrt[3]{2}+1)}=1+\frac{\sqrt[3]{2}-1}{(\sqrt[3]{2})^3-1}=1+\sqrt[3]{2}-1=\sqrt[3]{2}\)

c.

\(\sqrt{3}+\sqrt[3]{10+6\sqrt{3}}=\sqrt{3}+\sqrt[3]{(\sqrt{3}+1)^3}=\sqrt{3}+\sqrt{3}+1=2\sqrt{3}+1\)

d.

\(\frac{4+2\sqrt{3}}{\sqrt[3]{10+6\sqrt{3}}}=\frac{(\sqrt{3}+1)^2}{\sqrt[3]{(\sqrt{3}+1)^3}}=\frac{(\sqrt{3}+1)^2}{\sqrt{3}+1}=\sqrt{3}+1\)

e.

Đặt \(\sqrt[3]{2+10\sqrt{\frac{1}{27}}}=a; \sqrt[3]{2-10\sqrt{\frac{1}{27}}}=b\)

Khi đó:

$a^3+b^3=4$

$ab=\frac{2}{3}$

$E^3=(a+b)^3=a^3+b^3+3ab(a+b)$
$E^3=4+2E$

$E^3-2E-4=0$
$E^2(E-2)+2E(E-2)+2(E-2)=0$

$(E-2)(E^2+2E+2)=0$

Dễ thấy $E^2+2E+2>0$ nên $E-2=0$

$\Leftrightarrow E=2$

\(a,=\sqrt{6+2\sqrt{3-2\sqrt{3}+1}}\)

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

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

\(=\sqrt{4+2\sqrt{3}}\)

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

\(b,=\sqrt{6-2\sqrt{3+\sqrt{12+2\sqrt{12}+1}}}\)

\(=\sqrt{6-2\sqrt{3+\sqrt{12}+1}}\)

\(=\sqrt{6-2\sqrt{3+2\sqrt{3}+1}}\)

\(=\sqrt{6-2\left(\sqrt{3}+1\right)}=\sqrt{6-2\sqrt{3}-2}=\sqrt{4-2\sqrt{3}}\)

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

\(c,=\sqrt{\sqrt{3}+\sqrt{48-10\sqrt{4+2.2\sqrt{3}+3}}}\)

\(=\sqrt{\sqrt{3}+\sqrt{48-10\left(2+\sqrt{3}\right)}}\)

\(=\sqrt{\sqrt{3}+\sqrt{28-10\sqrt{3}}}\)

\(=\sqrt{\sqrt{3}+\sqrt{25-2.5\sqrt{3}+3}}\)

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

\(d,=\sqrt{23-6\sqrt{10+4\sqrt{2-2\sqrt{2}+1}}}\)

\(=\sqrt{23-6\sqrt{6+4\sqrt{2}}}\)

\(=\sqrt{23-6\sqrt{4+2.2\sqrt{2}+2}}\)

\(=\sqrt{23-6\sqrt{\left(2+\sqrt{2}\right)^2}}\)

\(=\sqrt{23-12-6\sqrt{2}}=\sqrt{11-6\sqrt{2}}\)

\(=\sqrt{9-2.3\sqrt{2}+2}=3-\sqrt{2}\)

a) Ta có: \(\sqrt{6+2\sqrt{4-2\sqrt{3}}}\)

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

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

b) Ta có: \(\sqrt{6-2\sqrt{3+\sqrt{13+4\sqrt{3}}}}\)

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

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

\(=\sqrt{4-2\sqrt{3}}=\sqrt{3}-1\)

c) Ta có: \(\sqrt{\sqrt{3}+\sqrt{48-10\sqrt{7+4\sqrt{3}}}}\)

\(=\sqrt{\sqrt{3}+\sqrt{48-10\left(2+\sqrt{3}\right)}}\)

\(=\sqrt{\sqrt{3}+\sqrt{28-10\sqrt{3}}}\)

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

\(=\sqrt{5}\)

d) Ta có: \(\sqrt{23-6\sqrt{10+4\sqrt{3-2\sqrt{2}}}}\)

\(=\sqrt{23-6\sqrt{10+4\left(\sqrt{2}-1\right)}}\)

\(=\sqrt{23-6\sqrt{6-4\sqrt{2}}}\)

\(=\sqrt{23-6\left(2-\sqrt{2}\right)}\)

\(=\sqrt{11+6\sqrt{2}}\)

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

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

2.\(\sqrt{10-4\sqrt{6}}=\sqrt{4-4\sqrt{6}+6}=\sqrt{\left(2-\sqrt{6}\right)^2}=\left|2-\sqrt{6}\right|=\sqrt{6}-2\)       \(\sqrt{15-6\sqrt{6}}=\sqrt{9+6\sqrt{6}+6}=\sqrt{\left(3+\sqrt{6}\right)^2}=3+\sqrt{6}\)

=>\(\sqrt{15-6\sqrt{6}}-\sqrt{10-4\sqrt{6}}\)=\(3+\sqrt{6}-\sqrt{6}+2\)=5

3. Tương tự bằng :\(8-3\sqrt{6}\)

1) \(\sqrt{6+2\sqrt{5}}\) = \(\sqrt{1+2.1.\sqrt{5}+\sqrt{5}^2}\) =  \(\sqrt{\left(1+\sqrt{5}\right)^2}\)

2) \(\sqrt{15-6\sqrt{6}}-\sqrt{10-4\sqrt{6}}\)

= \(\sqrt{3^2-2.3.\sqrt{6}+\sqrt{6}^2}\) - \(\sqrt{2^2.2.2.\sqrt{6}+\sqrt{6}^2}\)

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

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

= 3 + \(\sqrt{6}\) - 2 + \(\sqrt{6}\)

= 1 + 2\(\sqrt{6}\)

3) \(\sqrt{31-10\sqrt{6}}-\sqrt{\left(3-2\sqrt{6}\right)^2}\)

= \(\sqrt{5^2-2.5.\sqrt{6}+\sqrt{6}^2}\) - \(\sqrt{\left(3-2\sqrt{6}\right)^2}\)

= \(\sqrt{\left(5-\sqrt{6}\right)^2}\) - \(\sqrt{\left(3-2\sqrt{6}\right)^2}\)

= \(\left|5-\sqrt{6}\right|\) -  \(\left|3-2\sqrt{6}\right|\)

= 5 - \(\sqrt{6}-3-2\sqrt{6}\)

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

 Chúc bạn học tốt

\(\dfrac{2\sqrt{15}-2\sqrt{10}-3+\sqrt{6}}{2\sqrt{5}-2\sqrt{10}-\sqrt{3}+\sqrt{6}}\)

\(=\dfrac{2\sqrt{5}\left(\sqrt{3}-\sqrt{2}\right)-\sqrt{3}\left(\sqrt{3}-\sqrt{2}\right)}{2\sqrt{5}\left(1-\sqrt{2}\right)-\sqrt{3}\left(1-\sqrt{2}\right)}\)

\(=\dfrac{\sqrt{3}-\sqrt{2}}{1-\sqrt{2}}=-\sqrt{3}-\sqrt{6}+\sqrt{2}+2\)

\(\dfrac{6-\sqrt{6}}{\sqrt{6}-1}+\dfrac{6-\sqrt{6}}{\sqrt{6}}\)

\(=\dfrac{\sqrt{6}\cdot\sqrt{6}-\sqrt{6}}{\sqrt{6}-1}+\dfrac{\sqrt{6}\cdot\sqrt{6}-\sqrt{6}}{\sqrt{6}}\)

\(=\dfrac{\sqrt{6}\left(\sqrt{6}-1\right)}{\sqrt{6}-1}+\dfrac{\sqrt{6}\left(\sqrt{6}-1\right)}{\sqrt{6}}\)

\(=\dfrac{\sqrt{6}}{1}+\dfrac{\sqrt{6}-1}{1}\)

\(=\sqrt{6}+\sqrt{6}-1\)

\(=2\sqrt{6}-1\)

=======================

\(\dfrac{1}{\sqrt{2}-\sqrt{3}}-\dfrac{3}{\sqrt{18}+2\sqrt{3}}\)

\(=\dfrac{1}{\sqrt{2}-\sqrt{3}}-\dfrac{3}{\sqrt{6}\cdot\sqrt{3}+\sqrt{6}\cdot\sqrt{2}}\)

\(=\dfrac{1}{\sqrt{2}-\sqrt{3}}-\dfrac{3}{\sqrt{6}\left(\sqrt{3}+\sqrt{2}\right)}\)

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

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

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

\(=\dfrac{5\sqrt{3}}{-\sqrt{6}}=-\dfrac{5}{\sqrt{2}}\)

A = \(\sqrt[3]{6\sqrt{3}+10}-\sqrt[3]{6\sqrt{3}-10}=\sqrt[3]{\left(\sqrt{3}+1\right)^3}-\sqrt[3]{\left(\sqrt{3}-1\right)^3}=\sqrt{3}+1-\sqrt{3}+1=2\)

B = \(\dfrac{4+2\sqrt{3}}{\sqrt[3]{10+6\sqrt{3}}}=\dfrac{\left(\sqrt{3}+1\right)^2}{\sqrt{3}+1}=\sqrt{3}+1\)

C = \(\sqrt[4]{56-24\sqrt{5}}=\sqrt[4]{\left(6-\sqrt{20}\right)^2}=\sqrt[4]{\left(\sqrt{5}-1\right)^4}=\sqrt{5}-1\)