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

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

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

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

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

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

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

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

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

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

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

\(=2\)

a) \(=\sqrt{4+\sqrt{8}}.\sqrt{2-\sqrt{2}}=\sqrt{2\left(2+\sqrt{2}\right)\left(2-\sqrt{2}\right)}=\sqrt{2.2}=\sqrt{4}=2\)

b) \(=\sqrt{2}\left(\sqrt{5}+\sqrt{3}\right)\left(\sqrt{15}-4\right)\sqrt{4+\sqrt{15}}=\left(\sqrt{5}+\sqrt{3}\right)\left(\sqrt{15}-4\right)\sqrt{8+2\sqrt{15}}\)

\(=\left(\sqrt{5}+\sqrt{3}\right)\left(\sqrt{15}-4\right)\sqrt{\left(\sqrt{5}\right)^2+2.\sqrt{5}.\sqrt{3}+\left(\sqrt{3}\right)^2}\)

\(=\left(\sqrt{5}+\sqrt{3}\right)\left(\sqrt{15}-4\right)\sqrt{\left(\sqrt{5}+\sqrt{3}\right)^2}=\left(\sqrt{5}+\sqrt{3}\right)^2\left(\sqrt{15}-4\right)=2\left(4+\sqrt{15}\right)\left(\sqrt{15}-4\right)\)

\(=-2\)

ưu tiên phương pháp bình phương :

a) \(\left(4+\sqrt{15}\right)^2\left(\sqrt{10}-\sqrt{6}\right)^2\left(\sqrt{4-\sqrt{15}}\right)^2\)

\(=\left(4+\sqrt{15}\right)^2\left(4-\sqrt{15}\right)\left(\sqrt{10}-\sqrt{6}\right)^2\)

Tính ra kết quả nhớ căn đó

b) Phương pháp trục căn thức :

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

Trên tử có hàng đẳng thức . bạn tự quy động là ra 

mình vẫn chưa hiểu câu a

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

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

= \(\sqrt{8+2\sqrt{15}}\left(\sqrt{5}-\sqrt{3}\right)\sqrt{4^2-\left(\sqrt{15}\right)^2}\)

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

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

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

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

= \(5-3=2\)

a)   = \(\frac{\sqrt{2}.\sqrt{4+\sqrt{15}}}{\sqrt{2}}+\frac{\sqrt{2}.\sqrt{4-\sqrt{15}}}{\sqrt{2}}\)
      = \(\frac{\sqrt{8+2\sqrt{15}}}{\sqrt{2}}+\frac{\sqrt{8-2\sqrt{15}}}{\sqrt{2}}\)
      =  \(\frac{\sqrt{\left(\sqrt{5}+\sqrt{3}\right)^2}}{\sqrt{2}}+\frac{\sqrt{\left(\sqrt{5}-\sqrt{3}\right)^2}}{\sqrt{2}}\)
      = \(\frac{\sqrt{5}+\sqrt{3}}{\sqrt{2}}+\frac{\sqrt{5}-\sqrt{3}}{\sqrt{2}}\)

      =   \(\frac{2\sqrt{5}}{\sqrt{2}}\)
      =   \(\sqrt{10}\)
b) = \(5\sqrt{3}+4\sqrt{3}-10\sqrt{3}\)

    = \(-\sqrt{3}\)
c) = \(\sqrt{\left(\sqrt{2}-\sqrt{1}\right)^2}+\frac{1\left(\sqrt{2}+1\right)}{2-1}\)
    = \(\sqrt{2}-\sqrt{1}+\frac{1\left(\sqrt{2}+1\right)}{1}\)
    = \(\sqrt{2}-\sqrt{1}+\sqrt{2}+\sqrt{1}\)
    = \(2\sqrt{2}\)
Chúc bạn học tốt ^^

Tính

a) Ta có: \(A=\left(\sqrt{6}+\sqrt{10}\right)-\sqrt{4-\sqrt{15}}\)

\(=\sqrt{2}\left(\sqrt{3}+\sqrt{5}\right)-\sqrt{4-\sqrt{15}}\)

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

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

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

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

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

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

\(=2\sqrt{3}\)

c) Ta có: \(C=\left(4+\sqrt{15}\right)\left(\sqrt{10}-\sqrt{6}\right)\cdot\sqrt{4-\sqrt{15}}\)

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

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

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

\(=\left(4+\sqrt{15}\right)\cdot\left(\sqrt{5}-\sqrt{3}\right)\cdot\sqrt{\left(\sqrt{5}-\sqrt{3}\right)^2}\)

\(=\left(4+\sqrt{15}\right)\cdot\left(\sqrt{5}-\sqrt{3}\right)\cdot\left|\sqrt{5}-\sqrt{3}\right|\)

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

\(=\left(4+\sqrt{15}\right)\cdot\left(\sqrt{5}-\sqrt{3}\right)^2\)

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

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

\(=2\left[4^2-\left(\sqrt{15}\right)^2\right]\)

\(=2\cdot\left[16-15\right]=2\cdot1=2\)

`a)sqrt{4+sqrt7}-sqrt{4-sqrt7}`

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

`=sqrt{(7+2sqrt7+1)/2}-sqrt{(7-2sqrt7+1)/2}`

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

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

`=2/sqrt2=sqrt2`

`b)sqrt{4--sqrt15}-sqrt{4+sqrt15}`

`=sqrt{(8-2sqrt15)/2}-sqrt{(8+2sqrt15)/2}`

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

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

`=(sqrt5-sqrt3)/sqrt2-(sqrt5+sqrt3)/sqrt2`

`=(-2sqrt3)/sqrt2=-sqrt6`

`c)sqrt{2+sqrt3}+sqrt{2-sqrt3}`

`=sqrt{(4+2sqrt3)/2}+sqrt{(4-2sqrt3)/2}`

`=sqrt{(3+2sqrt3+1)/2}+sqrt{(3-2sqrt3+1)/2}`

`=sqrt{(sqrt3+1)^2/2}+sqrt{(sqrt3-1)^2/2}`

`=(sqrt3+1)/sqrt2+(sqrt3-1)/sqrt2`

`=(2sqrt3)/sqrt2=sqrt6`

`d)sqrt{9+sqrt17}-sqrt{9-sqrt17}`

`=sqrt{(18+2sqrt17)/2}-sqrt{(18-2sqrt17)/2}`

`=sqrt{(17+2sqrt17+1)/2}-sqrt{(17-2sqrt17+1)/2}`

`=sqrt{(sqrt17+1)^2/2}-sqrt{(sqrt17-1)^2/2}`

`=(sqrt17+1)/sqrt2-(sqrt17-1)/sqrt2`

`=2/sqrt2=sqrt2`

a: Ta có: \(\sqrt{4+\sqrt{7}}-\sqrt{4-\sqrt{7}}\)

\(=\dfrac{\sqrt{8+2\sqrt{7}}-\sqrt{8-2\sqrt{7}}}{\sqrt{2}}\)

\(=\dfrac{\sqrt{7}+1-\sqrt{7}+1}{\sqrt{2}}=\sqrt{2}\)

b: Ta có: \(\sqrt{4-\sqrt{15}}-\sqrt{4+\sqrt{15}}\)

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

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

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 \(D=\sqrt{2}\)

a: \(\sqrt{\left(4-\sqrt{15}\right)^2}+\sqrt{15}\)

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

b: \(\sqrt{7+4\sqrt{3}}-\sqrt{7-4\sqrt{3}}\)

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

\(=2\sqrt{3}\)

c: \(\sqrt{29+12\sqrt{5}}-\sqrt{29-12\sqrt{5}}\)

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

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