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, Sửa đề:

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

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

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

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

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

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

5: Ta có: \(\dfrac{2-\sqrt{2}}{1-\sqrt{2}}+\dfrac{\sqrt{2}-\sqrt{6}}{\sqrt{3}-1}\)

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

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

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

\(=\dfrac{1}{6\sqrt{6}}-\dfrac{6}{\sqrt{6}}=-\dfrac{35}{6\sqrt{6}}\)

b)\(\left(\sqrt{6}+\sqrt{5}\right)^2+\left(\sqrt{6}-\sqrt{5}\right)^2\)

\(=6+2\sqrt{30}+5+6-2\sqrt{30}+5=22\)

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

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

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

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

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

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

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

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

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

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

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

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

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

\(=\left(2\sqrt{5}+4-10-4\sqrt{5}+\sqrt{3}\right)\left(\sqrt{5}-\sqrt{3}\right)\)

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

\(=-20+2\sqrt{15}+\sqrt{15}-3-6\sqrt{5}+6\sqrt{3}\)

\(=-23+3\sqrt{15}-6\sqrt{5}+6\sqrt{3}\)

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

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

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

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

\(I=\left(\sqrt{2}-1\right).\left(\sqrt{2}+1\right)\)

\(I=2-1\)

I=1

g: \(=\left|\sqrt{6}-1\right|=\sqrt{6}-1\)

h: \(=\left|2\sqrt{3}-1\right|=2\sqrt{3}-1\)

l: \(=\left|2-\sqrt{3}\right|-2=2-\sqrt{3}-2=-\sqrt{3}\)

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

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