a) Ta có: \(A=\sqrt{\sqrt{3}+\sqrt{2}}\cdot\sqrt{\sqrt{3}-\sqrt{2}}\)

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

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

b) Ta có: \(B=\sqrt{5-2\sqrt{6}}+\sqrt{5+2\sqrt{6}}\)

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

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

\(=2\sqrt{3}\)

`A=sqrt{sqrt3+sqrt2}.sqrt{sqrt3-sqrt2}`

`=sqrt{(sqrt3+sqrt2)(sqrt3-sqrt2)}`

`=sqrt{3-2}=1`

`b)B=sqrt{5-2sqrt6}+sqrt{5+2sqrt6}`

`=sqrt{3-2sqrt6+2}+sqrt{3+2sqrt6+2}`

`=sqrt{(sqrt3-sqrt2)^2}+sqrt{(sqrt3+sqrt2)^2}`

`=sqrt3-sqrt2+sqrt3+sqrt2=2sqrt3`

`c)C=3-sqrt{3-sqrt5}`

`=3-sqrt{(6-2sqrt5)/2}`

`=3-sqrt{(sqrt5-1)^2/2}`

`=3-(sqrt5-1)/sqrt2`

`=3-(sqrt{10}-sqrt2)/2`

`=(6-sqrt{10}+sqrt2)/2`

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

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

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

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

\(=\dfrac{4\sqrt{3}-2\sqrt{3}}{2\sqrt{3}}\cdot\sqrt{2}=\sqrt{2}\)

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

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

=4

1:

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

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

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

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

\(a,=\dfrac{3-\sqrt{2}+3+\sqrt{2}}{\left(3+\sqrt{2}\right)\left(3-\sqrt{2}\right)}=\dfrac{6}{-1}=-6\\ b,=\dfrac{6\sqrt{2}+8-6\sqrt{2}+8}{\left(3\sqrt{2}-4\right)\left(3\sqrt{2}+4\right)}=\dfrac{16}{2}=8\\ c,=\dfrac{\left(\sqrt{5}-\sqrt{3}\right)^2+\left(\sqrt{5}+\sqrt{3}\right)^2}{\left(\sqrt{5}-\sqrt{3}\right)\left(\sqrt{5}+\sqrt{3}\right)}\\ =\dfrac{8-2\sqrt{15}+8+2\sqrt{15}}{2}=\dfrac{16}{2}=8\)

\(d,=\dfrac{6\sqrt{2}+9\sqrt{3}-6\sqrt{2}+9\sqrt{3}}{\left(2\sqrt{2}-3\sqrt{3}\right)\left(2\sqrt{2}+3\sqrt{3}\right)}=\dfrac{18\sqrt{3}}{-19}=\dfrac{-18\sqrt{3}}{19}\\ e,=\sqrt{\sqrt{5}-\sqrt{3-\sqrt{\left(2\sqrt{5}-3\right)^2}}}\\ =\sqrt{\sqrt{5}-\sqrt{6-2\sqrt{5}}}\\ =\sqrt{\sqrt{5}-\sqrt{\left(\sqrt{5}-1\right)^2}}\\ =\sqrt{\sqrt{5}-\sqrt{5}+1}=\sqrt{1}=1\)

a: Sửa đề: căn 6+2căn 5-căn 5

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

b: \(a^3=2-\sqrt{3}+2+\sqrt{3}+3a\)

=>a^3-3a-4=0

=>a^3-3a=4

\(\dfrac{64}{\left(a^2-3\right)^3}-3a=\left(\dfrac{4}{a^2-3}\right)^3-3a\)

\(=\left(\dfrac{a^3-3a}{a^2-3}\right)^3-3a=a^3-3a\)

=4

\(a,\dfrac{2}{\sqrt{3}-\sqrt{5}}+\dfrac{3-2\sqrt{3}}{\sqrt{3}-2}\)

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

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

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

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

\(b,\dfrac{5-\sqrt{5}}{\sqrt{5}-1}+\dfrac{\sqrt{5}-\sqrt{3}}{\sqrt{5}+3}\)

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

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

\(=\dfrac{2\sqrt{5}+8-2\sqrt{15}}{2}\)

\(=\dfrac{2\cdot\left(\sqrt{5}+4-\sqrt{15}\right)}{2}\)

\(=\sqrt{5}-\sqrt{15}+4\)

#\(Toru\)

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 $E-2=0$

$\Leftrightarrow E=2$