a)\(\left(\sqrt{21}+7\right)\cdot\sqrt{10-2\sqrt{21}}\)

\(=\left(\sqrt{21}+7\right)\sqrt{\left(\sqrt{7}-\sqrt{3}\right)^2}\)

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

\(=\sqrt{7}\left(7-3\right)=4\sqrt{7}\)

b)\(\left(7+\sqrt{14}\right)\sqrt{9-2\sqrt{14}}\)

\(=\left(7+\sqrt{14}\right)\sqrt{\left(\sqrt{7}-\sqrt{2}\right)^2}\)

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

\(=\sqrt{7}\left(7-2\right)=5\sqrt{7}\)

a) \(\left(\sqrt{14}+\sqrt{6}\right)\sqrt{5-\sqrt{21}}\)

\(=\sqrt{14}\cdot\sqrt{5-\sqrt{21}}+\sqrt{6}\cdot\sqrt{5-\sqrt{21}}\)

\(=\sqrt{14\cdot\left(5-\sqrt{21}\right)}+\sqrt{6\cdot\left(5-\sqrt{21}\right)}\)

\(=\sqrt{70-14\sqrt{21}}+\sqrt{30-6\sqrt{21}}\)

\(=\sqrt{7^2-2\cdot7\cdot\sqrt{21}+\left(\sqrt{21}\right)^2}+\sqrt{\left(\sqrt{21}\right)^2-2\cdot3\cdot\sqrt{21}+3^2}\)

\(=\sqrt{\left(7-\sqrt{21}\right)^2}+\sqrt{\left(\sqrt{21}-3\right)^2}\)

\(=\left|7-\sqrt{21}\right|+\left|\sqrt{21}-3\right|\)

\(=7-\sqrt{21}+\sqrt{21}-3\)

\(=4\)

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

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

\(=\left(4\sqrt{10}-4\sqrt{6}+\sqrt{150}-\sqrt{90}\right)\sqrt{4-\sqrt{15}}\)

\(=\left(4\sqrt{10}-4\sqrt{6}+5\sqrt{6}-3\sqrt{10}\right)\sqrt{4-\sqrt{15}}\)

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

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

\(=\sqrt{40-10\sqrt{15}}+\sqrt{24-6\sqrt{15}}\)

\(=\sqrt{5^2-2\cdot5\cdot\sqrt{15}+\left(\sqrt{15}\right)^2}+\sqrt{\left(\sqrt{15}\right)^2-2\cdot3\cdot\sqrt{15}+3^2}\)

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

\(=\left|5-\sqrt{15}\right|+\left|\sqrt{15}-3\right|\)

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

\(=2\)

Lời giải:
Đặt biểu thức là $A$. Ta có:

\(A=(5+\sqrt{21})(\sqrt{7}-\sqrt{3}).\sqrt{2}.\sqrt{5-\sqrt{21}}\)

\(=(5+\sqrt{21})(\sqrt{7}-\sqrt{3}).\sqrt{10-2\sqrt{21}}\)

\(=(5+\sqrt{21})(\sqrt{7}-\sqrt{3}).\sqrt{(\sqrt{7}-\sqrt{3})^2}\)

\(=(5+\sqrt{21})(\sqrt{7}-\sqrt{3})|\sqrt{7}-\sqrt{3}|=(5+\sqrt{21})(\sqrt{7}-\sqrt{3})^2\)

\(=(5+\sqrt{21})(10-2\sqrt{21})=2(5+\sqrt{21})(5-\sqrt{21})=2(5^2-21)=8\)

Ta có: \(\left(5+\sqrt{21}\right)\cdot\left(\sqrt{14}-\sqrt{6}\right)\cdot\sqrt{5-\sqrt{21}}\)

\(=\dfrac{\left(10+2\sqrt{21}\right)\cdot\left(\sqrt{7}-\sqrt{3}\right)\cdot\sqrt{10-2\sqrt{21}}}{2}\)

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

=8

`a)A=(3-sqrt5)sqrt{3+sqrt5}+(3+sqrt5)sqrt{3-sqrt5}`

`=sqrt{3-sqrt5}sqrt{3+sqrt5}(sqrt{3+sqrt5}+sqrt{3-sqrt5})`

`=sqrt{9-5}(sqrt{3+sqrt5}+sqrt{3-sqrt5})`

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

`=sqrt2(sqrt{6+2sqrt5}+sqrt{6-2sqrt5})`

`=sqrt2(sqrt{(sqrt5+1)^2}+sqrt{(sqrt5+1)^2})`

`=sqrt2(sqrt5+1+sqrt5-1)`

`=sqrt{2}.2sqrt5`

`=2sqrt{10}`

`b)B=(5+sqrt{21})(sqrt{14}-sqrt6)sqrt{5-sqrt{21}}`

`=sqrt{5+sqrt{21}}sqrt{5-sqrt{21}}sqrt{5+sqrt{21}}(sqrt{14}-sqrt6)`

`=sqrt{25-21}sqrt{5+sqrt{21}}(sqrt{14}-sqrt6)`

`=2sqrt{5+sqrt{21}}(sqrt{14}-sqrt6)`

`=2sqrt2sqrt{5+sqrt{21}}(sqrt{7}-sqrt3)`

`=2sqrt{10+2sqrt{21}}(sqrt{7}-sqrt3)`

`=2sqrt{(sqrt3+sqrt7)^2}(sqrt{7}-sqrt3)`

`=2(sqrt3+sqrt7)(sqrt{7}-sqrt3)`

`=2(7-3)`

`=8`

`c)C=sqrt{4+sqrt7}-sqrt{4-sqrt7}`

`=sqrt{(8+2sqrt7)/2}-sqrt{(8-2sqrt7)/2}`

`=sqrt{(sqrt7+1)^2/2}-sqrt{(sqrt7+1)^2/2}`

`=(sqrt7+1)/sqrt2-(sqrt7-1)/2`

`=2/sqrt2=sqrt2`

`d)D=\sqrt{2+sqrt3}+sqrt{14-5sqrt3}+sqrt2`

`=>sqrt2D=sqrt{4+2sqrt3}+sqrt{28-10sqrt3}+2`

`=>sqrt2D=sqrt{(sqrt3+1)^2}+sqrt{(5-sqrt3)^2}+2`

`=>sqrt2D=8`

`=>D=4sqrt2`

Bài 1

a) Đặt VT = A

<=> \(2\sqrt{2}A=\left(8+2\sqrt{15}\right)\left(\sqrt{10}-\sqrt{6}\right)\sqrt{8-2\sqrt{15}}\)

<=> \(2\sqrt{2}A=\left(\sqrt{5}+\sqrt{3}\right)^2.\sqrt{2}\left(\sqrt{5}-\sqrt{3}\right).\sqrt{\left(\sqrt{5}-\sqrt{3}\right)^2}\)

<=> \(2A=\left(\sqrt{5}+\sqrt{3}\right)^2.\left(\sqrt{5}-\sqrt{3}\right)^2\)

<=> 2A = \(\left(5-3\right)^2=4\)

<=> A = 2

b) Đặt VT = B

<=> \(2\sqrt{2}B=\left(10+2\sqrt{21}\right).\left(\sqrt{14}-\sqrt{6}\right)\sqrt{10-2\sqrt{21}}\)

<=> \(2\sqrt{2}B=\left(\sqrt{7}+\sqrt{3}\right)^2.\sqrt{2}\left(\sqrt{7}-\sqrt{3}\right).\sqrt{\left(\sqrt{7}-\sqrt{3}\right)^2}\)

<=> \(2B=\left(\sqrt{7}+\sqrt{3}\right)^2.\left(\sqrt{7}-\sqrt{3}\right)^2=\left(7-3\right)^2=16\)

<=> B = 8 

Bài 2

Đặt VT = A

<=> A² = \(\dfrac{\sqrt{5}+2+\sqrt{5}-2+2\sqrt{\left(\sqrt{5}+2\right)\left(\sqrt{5}-2\right)}}{2}\)

<=> A² = \(\dfrac{2\sqrt{5}+2\sqrt{5-4}}{2}=\dfrac{2\sqrt{5}+2}{2}=\sqrt{5}+1\)

<=> \(A=\sqrt{\sqrt{5}+1}\)

\(\left(5+\sqrt{21}\right)\left(\sqrt{14}-\sqrt{6}\right)\sqrt{5-\sqrt{21}}=\left(5+\sqrt{21}\right)\left(\sqrt{7}-\sqrt{3}\right)\sqrt{7-2\sqrt{7}.\sqrt{3}+3}=\left(5+\sqrt{21}\right)\left(\sqrt{7}-\sqrt{3}\right)^2=\left(5+\sqrt{21}\right)\left(10-2\sqrt{21}\right)=2\left(5+\sqrt{21}\right)\left(5-\sqrt{21}\right)=2\left(25-21\right)=2.4=8\)

b: \(=\left(5+\sqrt{21}\right)\left(\sqrt{7}-\sqrt{3}\right)\sqrt{10-2\sqrt{21}}\)

\(=\left(5+\sqrt{21}\right)\left(10-2\sqrt{21}\right)\)

\(=50-10\sqrt{21}+10\sqrt{21}-42=8\)

a: \(A=\sqrt{\sqrt{2}-1}+\sqrt{\sqrt{2}+1}\)

=>\(A^2=\sqrt{2}-1+\sqrt{2}+1+2\sqrt{2-1}=2\sqrt{2}+2\)

=>\(A=\sqrt{2\sqrt{2}+2}\)

Đặt \(B=\sqrt{\sqrt{2}-1}+\sqrt{\sqrt{2}+1}-\sqrt{2+\sqrt{2}}\)

=>\(B=\sqrt{2\sqrt{2}+2}-\sqrt{2+\sqrt{2}}\)

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

=>\(B^2=4+3\sqrt{2}-2\sqrt[4]{2}\left(2+\sqrt{2}\right)\)

=>\(B\simeq0,35\)

\(A=\)bn ghi lại đề nha mình lười

\(=\left(\sqrt{5+\sqrt{21}}\right)^2\left(\sqrt{14}-\sqrt{6}\right)\sqrt{5-\sqrt{21}}\)

\(=\left(\sqrt{5+\sqrt{21}}\right)\left(\sqrt{5-\sqrt{21}}\right)\left(\sqrt{5+\sqrt{21}}\right)\left(\sqrt{14}-\sqrt{6}\right)\)

\(=\left(\sqrt{\left(5^2-21\right)}\right)\left(\sqrt{5+\sqrt{21}}\right)\left(\sqrt{14}-\sqrt{6}\right)\)

\(=2.\left(\sqrt{5+\sqrt{21}}\right)\sqrt{2}.\left(\sqrt{7}-\sqrt{3}\right)\)

\(=2.\left(\sqrt{10+2\sqrt{21}}\right)\left(\sqrt{7}-\sqrt{3}\right)\)

\(=2.\left(\sqrt{7+2\sqrt{21}+3}\right) \left(\sqrt{7}-\sqrt{3}\right)\)

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

\(=2.\left(\sqrt{7}+\sqrt{3}\right)\left(\sqrt{7}-\sqrt{3}\right)=2.\left(7-3\right)=2.4=8\)

tíck mình nha bn thanks nhìu !!!!!!!!!

\(\left(5+\sqrt{21}\right)\left(\sqrt{14}-\sqrt{6}\right)\sqrt{5-\sqrt{21}}=\left(5+\sqrt{21}\right)\left(\sqrt{7}-\sqrt{3}\right)\sqrt{10-2\sqrt{21}}=\left(5+\sqrt{21}\right)\left(\sqrt{7}-\sqrt{3}\right)\sqrt{\left(\sqrt{3}\right)^2-2.\sqrt{3}.\sqrt{7}+\left(\sqrt{7}\right)^2}=\left(5+\sqrt{21}\right)\left(\sqrt{7}-\sqrt{3}\right)\sqrt{\left(\sqrt{3}-\sqrt{7}\right)^2}=\left(5+\sqrt{21}\right)\left(\sqrt{7}-\sqrt{3}\right)^2=\left(5+\sqrt{21}\right)\left(10-2\sqrt{21}\right)=2\left(5+\sqrt{21}\right)\left(5-\sqrt{21}\right)=2\left(25-21\right)=2\cdot4=8\)