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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

\(=\dfrac{4+3\sqrt{2}+2\sqrt{2}-3}{-1}=-1-5\sqrt{2}\)

b: \(=\dfrac{1}{\sqrt{10}+\sqrt{6}}-\dfrac{1}{\sqrt{10}-\sqrt{6}}\)

\(=\dfrac{\sqrt{10}-\sqrt{6}-\sqrt{10}-\sqrt{6}}{4}=\dfrac{-2\sqrt{6}}{4}=-\dfrac{\sqrt{6}}{2}\)

c: \(\dfrac{-2}{3\sqrt{8}}+\dfrac{1}{3-2\sqrt{2}}\)

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

\(=\dfrac{10\sqrt{2}-6}{6\sqrt{2}\left(3-2\sqrt{2}\right)}=\dfrac{10-3\sqrt{2}}{6\left(3-2\sqrt{2}\right)}=\dfrac{18+11\sqrt{2}}{6}\)

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 $E-2=0$

$\Leftrightarrow E=2$

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\(2\sqrt{40\sqrt{3}}-2\sqrt{\sqrt{75}}-3\sqrt{5\sqrt{48}}\)

\(=2\cdot\sqrt{40\sqrt{3}}-2\cdot\sqrt{5\sqrt{3}}-3\cdot\sqrt{20\sqrt{3}}\)

\(=2\cdot2\sqrt{10}\cdot\sqrt{\sqrt{3}}-2\cdot\sqrt{5}\cdot\sqrt{\sqrt{3}}-6\sqrt{5}\cdot\sqrt{\sqrt{3}}\)

\(=4\sqrt{10}\sqrt{\sqrt{3}}-4\cdot\sqrt{5}\cdot\sqrt{\sqrt{3}}\)

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

2: \(=\dfrac{6\left(1-\sqrt{3}\right)}{1-\sqrt{3}}+\dfrac{3\left(\sqrt{3}+1\right)}{\sqrt{3}+1}=6+3=9\)

3: \(=\sqrt{3}+1-\sqrt{3}=1\)

đề là rút gọn các biểu thức sau

nhờ mọi người giải giúp mình. cảm ơn mn nhìu

a: \(=\dfrac{\sqrt{6-2\sqrt{5}}\left(3+\sqrt{5}\right)}{2\left(\sqrt{5}+1\right)}\)

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

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

b: \(=\dfrac{2\sqrt{5}\left(\sqrt{5}+\sqrt{2}\right)}{\sqrt{5}+\sqrt{2}}-2-2\sqrt{5}\)

=2căn 5-2-2căn 5

=-2

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

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

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