`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: \(\sqrt{4-\sqrt{15}}-\sqrt{4+\sqrt{15}}\)

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

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

\(a,A=\left(\dfrac{x+14\sqrt{x}-5}{x-25}+\dfrac{\sqrt{x}}{\sqrt{x}+5}\right):\dfrac{\sqrt{x}+2}{\sqrt{x}-5}\)

\(\Rightarrow A=\left(\dfrac{x+14\sqrt{x}-5}{\left(\sqrt{x}+5\right)\left(\sqrt{x}-5\right)}+\dfrac{\sqrt{x}\left(\sqrt{x}-5\right)}{\left(\sqrt{x}+5\right)\left(\sqrt{x}-5\right)}\right).\dfrac{\sqrt{x}-5}{\sqrt{x}+2}\)

\(\Rightarrow A=\left(\dfrac{x+14\sqrt{x}-5}{\left(\sqrt{x}+5\right)\left(\sqrt{x}-5\right)}+\dfrac{x-5\sqrt{x}}{\left(\sqrt{x}+5\right)\left(\sqrt{x}-5\right)}\right).\dfrac{\sqrt{x}-5}{\sqrt{x}+2}\)

\(\Rightarrow A=\dfrac{x+14\sqrt{x}-5+x-5\sqrt{x}}{\left(\sqrt{x}+5\right)\left(\sqrt{x}-5\right)}.\dfrac{\sqrt{x}-5}{\sqrt{x}+2}\)

\(\Rightarrow A=\dfrac{2x+9\sqrt{x}-5}{\left(\sqrt{x}+5\right)\left(\sqrt{x}-5\right)}.\dfrac{\sqrt{x}-5}{\sqrt{x}+2}\)

\(\Rightarrow A=\dfrac{2x+10\sqrt{x}-\sqrt{x}-5}{\left(\sqrt{x}+5\right)\left(\sqrt{x}+2\right)}\)

\(\Rightarrow A=\dfrac{2\sqrt{x}\left(\sqrt{x}+5\right)-\left(\sqrt{x}+5\right)}{\left(\sqrt{x}+5\right)\left(\sqrt{x}+2\right)}\)

\(\Rightarrow A=\dfrac{\left(2\sqrt{x}-1\right)\left(\sqrt{x}+5\right)}{\left(\sqrt{x}+5\right)\left(\sqrt{x}+2\right)}\)

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

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

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

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

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

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

a) \(\sqrt{3+2\sqrt{2}}-\sqrt{17-12\sqrt{2}}\)

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

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

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

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

b) \(\sqrt{5-2\sqrt{6}}-\sqrt{14-4\sqrt{6}}-\sqrt{48}\)

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

= \(\left|\sqrt{3}-\sqrt{2}\right|-\left|2\sqrt{3}-\sqrt{2}\right|-4\sqrt{3}\)

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

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

c) \(\sqrt{11+3\sqrt{8}}-\sqrt{17-12\sqrt{2}}-4\sqrt{8}\)

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

= \(\left|3+\sqrt{2}\right|-\left|3-2\sqrt{2}\right|-8\sqrt{2}\)

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

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

Cho mình sửa đề xí ạ! 

b) \(\frac{\sqrt{10}+\sqrt{15}}{\sqrt{8}+\sqrt{12}}\)

băng 3 bạn ạ

\(\sqrt{9-4\sqrt{5}}\)

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

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

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

\(\sqrt{16-2\sqrt{55}}\)

=\(\sqrt{11-2\sqrt{11}.\sqrt{5}+5}\)

=\(\sqrt{\left(\sqrt{11}-\sqrt{5}\right)^2}\)

=\(\sqrt{11}-\sqrt{5}\)

\(\sqrt{14+2\sqrt{33}}\)

=\(\sqrt{11+2\sqrt{11}.\sqrt{3}+3}\)

=\(\sqrt{\left(\sqrt{11}+\sqrt{3}\right)^2}\)

=\(\sqrt{11}+\sqrt{3}\)