a: \(A=\left(4+\sqrt{15}\right)\cdot\left(\sqrt{5}-\sqrt{3}\right)\cdot\sqrt{8-2\sqrt{15}}\)

\(=\left(4+\sqrt{15}\right)\left(8-2\sqrt{15}\right)\)

\(=32-8\sqrt{15}+8\sqrt{15}-30=2\)

b: \(\sqrt{2}\cdot B=\left(3-\sqrt{5}\right)\left(\sqrt{5}+1\right)+\left(3+\sqrt{5}\right)\left(\sqrt{5}-1\right)\)

\(\Leftrightarrow B\sqrt{2}=3\sqrt{5}+3-5-\sqrt{5}+3\sqrt{5}-3+5-\sqrt{5}\)

\(\Leftrightarrow B\sqrt{2}=4\sqrt{5}\)

hay \(B=2\sqrt{10}\)

d: \(D\sqrt{2}=\sqrt{5}+\sqrt{3}+\sqrt{5}-\sqrt{3}-2\cdot\left(\sqrt{5}-1\right)\)

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

hay \(D=\sqrt{2}\)

`1)A=sqrt{4+sqrt{10+2sqrt5}}+sqrt{4-sqrt{10+2sqrt5}}`

`<=>A^2=4+sqrt{10+2sqrt5}+4-sqrt{10+2sqrt5}+2sqrt{16-10-2sqrt5}`

`<=>A^2=8+2sqrt{6-2sqrt5}`

`<=>A^2=8+2sqrt{(sqrt5-1)^2}`

`<=>A^2=8+2(sqrt5-1)`

`<=>A^2=6+2sqrt5=(sqrt5+1)^2`

`<=>A=sqrt5+1(do \ A>0)`

`b)B=sqrt{35+12sqrt6}-sqrt{35-12sqrt6}`

Vì `35+12sqrt6>35-12sqrt6`

`=>B>0`

`B^2=35+12sqrt6+35-12sqrt6-2sqrt{35^2-(12sqrt6)^2}`

`<=>B^2=70-2sqrt{361}`

`<=>B^2=70-2sqrt{19^2}=70-38=32`

`<=>B=sqrt{32}=4sqrt2(do \ B>0)`

`3)(4+sqrt{15})(sqrt{10}-sqrt6)sqrt{4-sqrt{15}}`

`=sqrt{4+sqrt{15}}.sqrt{4-sqrt{15}}.sqrt{4+sqrt{15}}(sqrt{10}-sqrt6)`

`=sqrt{16-15}.sqrt2(sqrt5-sqrt3).sqrt{4+sqrt{15}}`

`=sqrt{8+2sqrt{15}}(sqrt5-sqrt3)`

`=sqrt{5+2sqrt{5.3}+3}(sqrt5-sqrt3)`

`=sqrt{(sqrt5+sqrt3)^2}(sqrt5-sqrt3)`

`=(sqrt5+sqrt3)(sqrt5-sqrt3)`

`=5-3=2`

1.\(\left(\sqrt{2}+1\right)^3-\left(\sqrt{2}-1\right)^3=2\sqrt{2}+6+3\sqrt{2}+1-\left(2\sqrt{2}-6+3\sqrt{2}-1\right)=14\)

2.\(\sqrt{4-\sqrt{15}}+\sqrt{4+\sqrt{15}}-2\sqrt{3-\sqrt{5}}\)

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

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

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

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

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

3.\(\dfrac{10+2\sqrt{10}}{\sqrt{5}+\sqrt{2}}+\dfrac{8}{1-\sqrt{5}}=\dfrac{\sqrt{20}\left(\sqrt{5}+\sqrt{2}\right)}{\sqrt{5}+\sqrt{2}}+\dfrac{8\left(1+\sqrt{5}\right)}{1-\left(\sqrt{5}\right)^2}\)

\(=\sqrt{20}+\dfrac{8\left(1+\sqrt{5}\right)}{-4}=2\sqrt{5}-2\left(1+\sqrt{5}\right)=-2\)

4.\(\sqrt{\dfrac{2-\sqrt{3}}{2+\sqrt{3}}}+\sqrt{\dfrac{2+\sqrt{3}}{2-\sqrt{3}}}\)

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

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

\(=\dfrac{\left(\sqrt{3}-1\right)^2+\left(\sqrt{3}+1\right)^2}{\left(\sqrt{3}+1\right)\left(\sqrt{3}-1\right)}=\dfrac{8}{3-1}=4\)

3: Ta có: \(\dfrac{10+2\sqrt{10}}{\sqrt{5}+\sqrt{2}}+\dfrac{8}{1-\sqrt{5}}\)

\(=\dfrac{2\sqrt{5}\left(\sqrt{5}+\sqrt{2}\right)}{\sqrt{5}+\sqrt{2}}-\dfrac{8\left(\sqrt{5}+1\right)}{\left(\sqrt{5}-1\right)\left(\sqrt{5}+1\right)}\)

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

=-2

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

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

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

=4

2: \(\dfrac{\sqrt{108}}{\sqrt{3}}=6\)

13: \(\sqrt{8-2\sqrt{15}}-\sqrt{23-4\sqrt{15}}\)

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

\(=-\sqrt{5}\)

14: \(\left(4+\sqrt{15}\right)\cdot\left(\sqrt{10}-\sqrt{6}\right)\cdot\sqrt{4-\sqrt{15}}\)

\(=\left(4+\sqrt{15}\right)\left(8-2\sqrt{15}\right)\)

=2

12.

\(\dfrac{\sqrt{108}}{\sqrt{3}}=\dfrac{\sqrt{36}.\sqrt{3}}{\sqrt{3}}=\sqrt{36}=6\)

13.

\(\sqrt{8-2\sqrt{15}}-\sqrt{23-4\sqrt{15}}\)

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

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

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

\(=-\sqrt{5}\)

14.

\(\left(4+\sqrt{15}\right)\left(\sqrt{10}-\sqrt{6}\right)\sqrt{4-\sqrt{15}}\)

\(=\sqrt{4+\sqrt{15}}\left(\sqrt{10}-\sqrt{6}\right)\sqrt{\left(4-\sqrt{15}\right)\left(4+\sqrt{15}\right)}\)

\(=\sqrt{8+2\sqrt{15}}\left(\sqrt{5}-\sqrt{3}\right)\sqrt{16-15}\)

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

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

\(=2\)

\(\sqrt{10+2\sqrt{10}+2\sqrt{15}+2\sqrt{6}}\\ =\sqrt{\sqrt{10}.\sqrt{10}+2\sqrt{10}+2\sqrt{3}.\sqrt{5}+2\sqrt{3}.\sqrt{2}}\\ =\sqrt{\sqrt{10}\left(\sqrt{10}+2\right)+\sqrt{3}\left(2\sqrt{5}+2\sqrt{2}\right)}\\ =\sqrt{\sqrt{10}\left[\sqrt{2}\left(\sqrt{5}+\sqrt{2}\right)\right]+\sqrt{3}\left[2\left(\sqrt{5}+\sqrt{2}\right)\right]}\)

\(=\sqrt{\sqrt{20}\left(\sqrt{5}+\sqrt{2}\right)+2\sqrt{3}\left(\sqrt{5}+\sqrt{2}\right)}\\ =\sqrt{\left(\sqrt{5}+\sqrt{2}\right)\left(2\sqrt{5}+2\sqrt{3}\right)}\)

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

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

\(\left(3-\sqrt{2}\right)\sqrt{7+4\sqrt{3}}\\ =\left(3-\sqrt{2}\right)\sqrt{\sqrt{3^2}+2.2\sqrt{3}+2^2}\\ =\left(3-\sqrt{2}\right)\sqrt{\left(\sqrt{3}+2\right)^2}\\ =\left(3-\sqrt{2}\right)\left|\sqrt{3}+2\right|\\ =\left(3-\sqrt{2}\right)\left(\sqrt{3}+2\right)\\ =3\sqrt{3}+6-\sqrt{6}-2\sqrt{2}\)

a, Dễ thấy C>0.

Ta có: \(C^2=\left(\sqrt{4+\sqrt{10+2\sqrt{5}}}+\sqrt{4-\sqrt{10+2\sqrt{5}}}\right)^2=4+\sqrt{10+2\sqrt{5}}+4-\sqrt{10+2\sqrt{5}}+2\sqrt{\left(4+\sqrt{10+2\sqrt{5}}\right)\left(4-\sqrt{10+2\sqrt{5}}\right)}=8+2\sqrt{16-10-2\sqrt{5}}=8+2\sqrt{6-2\sqrt{5}}=8+2\sqrt{\left(\sqrt{5}-1\right)^2}=8+2\left(\sqrt{5}-1\right)=8+2\sqrt{5}-2=6+2\sqrt{5}=\left(\sqrt{5}+1\right)^2\)

=>\(C=\sqrt{\left(\sqrt{5}+1\right)^2}=\left|\sqrt{5}+1\right|=\sqrt{5}+1\)(vì C>0).

Ta có: \(C=\dfrac{\sqrt{2}+\sqrt{3}+\sqrt{6}+\sqrt{8}+4}{\sqrt{2}+\sqrt{3}+\sqrt{4}}\)

\(=\dfrac{\sqrt{2}+\sqrt{3}+\sqrt{4}+\sqrt{4}+\sqrt{6}+\sqrt{8}}{\sqrt{2}+\sqrt{3}+\sqrt{4}}\)

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

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

Ta có: \(B=\dfrac{\sqrt{2-\sqrt{3}}+\sqrt{4-\sqrt{15}}+\sqrt{10}}{\sqrt{23-3\sqrt{5}}}\)

\(=\dfrac{\sqrt{4-2\sqrt{3}}+\sqrt{8-2\sqrt{15}}+2\sqrt{5}}{3\sqrt{5}-1}\)

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

=1

1/ Tính: \(A=\dfrac{\sqrt{15-10\sqrt{2}}+\sqrt{13+4\sqrt{10}}-\sqrt{11+2\sqrt{10}}}{2\sqrt{3+2\sqrt{2}}+\sqrt{9-4\sqrt{2}}+\sqrt{12+8\sqrt{2}}}=\dfrac{\sqrt{\left(\sqrt{10}-\sqrt{5}\right)^2}+\sqrt{\left(2\sqrt{2}+\sqrt{5}\right)^2}-\sqrt{\left(\sqrt{10}+1\right)^2}}{2\sqrt{\left(\sqrt{2}+1\right)^2}+\sqrt{\left(2\sqrt{2}-1\right)^2}+\sqrt{\left(2\sqrt{2}+2\right)^2}}=\dfrac{\sqrt{10}-\sqrt{5}+2\sqrt{2}+\sqrt{5}-\sqrt{10}-1}{2\sqrt{2}+2+2\sqrt{2}-1+2\sqrt{2}+2}=\dfrac{2\sqrt{2}-1}{6\sqrt{2}-3}=\dfrac{2\sqrt{2}-1}{3\left(2\sqrt{2}-1\right)}=\dfrac{1}{3}\)

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