\(\dfrac{\sqrt{5}\left(\sqrt{5}-2\right)}{\sqrt{5}-2}-\dfrac{11\left(4-\sqrt{5}\right)}{16-5}=\sqrt{5}-4+\sqrt{5}=2\sqrt{5}-4\)

\(=\sqrt{5}-4+\sqrt{5}=2\sqrt{5}-4\)

`a)\sqrt{9-4sqrt5}-sqrt5`

`=sqrt{5-2.2sqrt5+4}-sqrt5`

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

`=|\sqrt5-2|-sqrt5`

`=sqrt5-2-sqrt5=-2`

`b)\sqrt{7-4sqrt3}+sqrt{4-2sqrt3}`

`=\sqrt{4-2.2sqrt3+3}+\sqrt{3-2sqrt3+1}`

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

`=|2-sqrt3|+|sqrt3-1|`

`=2-sqrt3+sqrt3-1=1`

`c)(x-49)/(sqrtx-7)(x>=0,x ne 49)`

`=((sqrtx-7)(sqrtx+7))/(sqrtx-7)`

`=sqrtx+7`

`d)\sqrt{4+2\sqrt3}-\sqrt{13+4sqrt3}`

`=\sqrt{3+2sqrt3+1}-\sqrt{12+2.2sqrt3+1}`

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

`=sqrt3+1-2sqrt3-1=-sqrt3`

`e)2+sqrt{17-4sqrt{9+4sqrt{45}}}`(câu này hơi sai)

a) Ta có: \(9+4\sqrt{5}\)

\(=5+2\cdot\sqrt{5}\cdot2+4\)

\(=\left(\sqrt{5}+2\right)^2\)(đpcm)

b) Ta có: \(\sqrt{9-4\sqrt{5}}-\sqrt{5}\)

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

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

=-2(ddpcm)

c) Ta có: \(\left(4-\sqrt{7}\right)^2\)

\(=16-2\cdot4\cdot\sqrt{7}+7\)

\(=23-8\sqrt{7}\)(đpcm)

d) Ta có: \(\sqrt{17-12\sqrt{2}}+2\sqrt{2}\)

\(=\sqrt{9-2\cdot3\cdot2\sqrt{2}+8}+2\sqrt{2}\)

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

\(=3-2\sqrt{2}+2\sqrt{2}=3\)(đpcm)

\(a.VT=4+4\sqrt{5}+5=2^2+4\sqrt{5}+\sqrt{5}^2=\left(2+\sqrt{5}\right)^2=VP\)

\(b.\) Dựa vào câu a ta có: \(9-4\sqrt{5}=\left(\sqrt{5}-2\right)^2\)

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

\(c.VT=16-8\sqrt{7}+7=4^2-8\sqrt{7}+\sqrt{7}^2=\left(4-\sqrt{7}\right)^2=VP\)

\(d.\) 

Ta có: \(17-12\sqrt{2}=8-12\sqrt{2}+9=\left(2\sqrt{2}\right)^2-12\sqrt{2}+3^2=\left(2\sqrt{2}-3\right)^2\)

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

Bài làm của: Phùng Khánh Linh

c)\(\sqrt{17-12\sqrt{2}}-\sqrt{24-8\sqrt{8}}\)

= \(\sqrt{3^2-2.3.2\sqrt{2}+\left(2\sqrt{2}\right)^2}\) \(-\) \(\sqrt{4^2-2.4.\sqrt{8}+\left(\sqrt{8}\right)^2}\)

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

= \(\left|3-2\sqrt{2}\right|-\left|4-\sqrt{8}\right|\)

= (3 - 2\(\sqrt{2}\)) - (4 - \(\sqrt{8}\))

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

= -1

\(a.\sqrt{4+2\sqrt{3}}-\sqrt{4-2\sqrt{3}}=\sqrt{3+2\sqrt{3}.1+1}-\sqrt{3-2\sqrt{3}.1+1}=\sqrt{\left(\sqrt{3}+1\right)^2}-\sqrt{\left(\sqrt{3}-1\right)^2}=\text{|}\sqrt{3}+1\text{|}-\text{|}\sqrt{3}-1\text{|}=2\)\(b.\sqrt{9-4\sqrt{5}}-\sqrt{9+4\sqrt{5}}=\sqrt{5-4\sqrt{5}+4}-\sqrt{5+4\sqrt{5}+4}=\sqrt{\left(\sqrt{5}-2\right)^2}-\sqrt{\left(\sqrt{5}+2\right)^2}=\text{|}\sqrt{5}-2\text{|}-\text{|}\sqrt{5}+2\text{|}=-4\) Còn lại tương tự nhé .

Ta có : \(\sqrt{17-4\sqrt{9+4\sqrt{5}}}\)\(=\sqrt{17-4\sqrt{4+2.2\sqrt{5}+5}}\)

\(=\sqrt{17-4\sqrt{\left(2+\sqrt{5}\right)^2}}=\sqrt{17-4\left(2+\sqrt{5}\right)}=\sqrt{17-8-4\sqrt{5}}\)

\(=\sqrt{9-4\sqrt{5}}=\sqrt{4-2.2\sqrt{5}+5}=\sqrt{\left(2-\sqrt{5}\right)^2}=\sqrt{5}-2\)

`sqrt{17-4sqrt{9+4sqrt5}}`

`=sqrt{17-4sqrt{5+2.2.sqrt5+4}}`

`=sqrt{17-4sqrt{(2+sqrt5)^2}}`

`=sqrt{17-4(2+sqrt5)}`

`=sqrt{17-8-4sqrt5}`

`=sqrt{9-4sqrt5}`

`=sqrt{5-2.2.sqrt5+4}`

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

`=|sqrt5-2|`

`=sqrt5-2`

\(\sqrt{10+2\sqrt{17-4\sqrt{9+4\sqrt{5}}}}\)

\(=\sqrt{10+2\sqrt{17-4\sqrt{2^2+2\cdot2\sqrt{5}+\left(\sqrt{5}\right)^2}}}\)

\(=\sqrt{10+2\sqrt{17-4\sqrt{\left(2+\sqrt{5}\right)^2}}}\)

\(=\sqrt{10+2\sqrt{17-4\cdot\left|2+\sqrt{5}\right|}}\)

\(=\sqrt{10+2\sqrt{17-4\cdot\left(2+\sqrt{5}\right)}}\)

\(=\sqrt{10+2\sqrt{17-8-4\sqrt{5}}}\)

\(=\sqrt{10+2\sqrt{9-4\sqrt{5}}}\)

\(=\sqrt{10+2\sqrt{2^2-2\cdot2\cdot\sqrt{5}+\left(\sqrt{5}\right)^2}}\)

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

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

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

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

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

\(=\sqrt{\left(\sqrt{5}\right)^2+2\sqrt{5}\cdot1+1^2}\)

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

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

\(=\sqrt{5}+1\)

\(=\sqrt{10+2\sqrt{17-4\sqrt{5+4\sqrt{5}+4}}}\)

\(=\sqrt{10+2\sqrt{17-4\sqrt{\left(\sqrt{5}+2\right)^2}}}\)

\(=\sqrt{10+2\sqrt{17-4\cdot\left|\sqrt{5}+2\right|}}\)

\(=\sqrt{10+2\sqrt{17-4\left(\sqrt{5}+2\right)}}\) (vì \(\sqrt{5}+2>0\))

\(=\sqrt{10+2\sqrt{17-4\sqrt{5}-8}}\)

\(=\sqrt{10+2\sqrt{9-4\sqrt{5}}}\\ =\sqrt{10+2\sqrt{5-4\sqrt{5}+4}}\\ =\sqrt{10+2\sqrt{\left(\sqrt{5}-2\right)^2}}\\ =\sqrt{10+2\cdot\left|\sqrt{5}-2\right|}\)

\(=\sqrt{10+2\cdot\left(\sqrt{5}-2\right)}\) (vì \(\sqrt{5}-2>0\))

\(=\sqrt{10+2\sqrt{5}-4}\\ =\sqrt{6+2\sqrt{5}}\\ =\sqrt{5+2\sqrt{5}+1}\\ =\sqrt{\left(\sqrt{5}+1\right)^2}\\ =\left|\sqrt{5}+1\right|\)

\(=\sqrt{5}+1\) (vì \(\sqrt{5}+1>0\))

\(1.\sqrt{17-4\sqrt{9+4\sqrt{5}}}=\sqrt{17-4\sqrt{5+2.2\sqrt{5}+4}}=\sqrt{17-4\left(\sqrt{5}+2\right)}=\sqrt{5-2.2\sqrt{5}+4}=\sqrt{5}-2\)

\(2.\sqrt{17-6\sqrt{2+\sqrt{9+4\sqrt{2}}}}=\sqrt{17-6\sqrt{2+\sqrt{8+2.2\sqrt{2}+1}}}=\sqrt{17-6\sqrt{2+2\sqrt{2}+1}}=\sqrt{17-6\left(\sqrt{2}+1\right)}=\sqrt{9-2.3\sqrt{2}+2}=3-\sqrt{2}\)\(3.\sqrt{3+\sqrt{5-\sqrt{13+4\sqrt{3}}}}=\sqrt{3+\sqrt{5-\sqrt{12+2.2\sqrt{3}+1}}}=\sqrt{3+\sqrt{3-2\sqrt{3}+1}}=\sqrt{2+\sqrt{3}}=\dfrac{\sqrt{3+2\sqrt{3}+1}}{\sqrt{2}}=\dfrac{\sqrt{3}+1}{\sqrt{2}}\)

\(4.\sqrt{27+10\sqrt{2}}:\dfrac{1}{\sqrt{\left(\sqrt{2}-5\right)^2}}=\sqrt{25+2.5\sqrt{2}+2}.\left(5-\sqrt{2}\right)=\left(5+\sqrt{2}\right)\left(5-\sqrt{2}\right)=5-2=3\)

a) \(\sqrt{24+8\sqrt{5}}+\sqrt{9-4\sqrt{5}}\)

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

\(=3\sqrt{5}\)

b) \(\sqrt{17-12\sqrt{2}}+\sqrt{9+4\sqrt{2}}\)

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

=2

c) \(\sqrt{6-4\sqrt{2}}+\sqrt{22-12\sqrt{2}}\)

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

\(=2\sqrt{2}\)

`2+\sqrt{17-4sqrt{9+4sqrt5}}`

`=2+sqrt{17-4sqrt{4+2.2sqrt5+5}}`

`=2+sqrt{17-4sqrt{(sqrt5+2)^2}}`

`=2+sqrt{17-4(sqrt5+2)}`

`=2+sqrt{9-4sqrt5}`

`=2+sqrt{5-2.2sqrt5+4}`

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

`=2+sqrt5-2=sqrt5`

a.\(\sqrt{7+4\sqrt{3}}=\sqrt{\left(\sqrt{3}+2\right)^2}=\left|\sqrt{3}+2\right|=\sqrt{3}+2\)

b.\(\sqrt{9-4\sqrt{5}}=\sqrt{\left(\sqrt{5}-2\right)^2}=\left|\sqrt{5}-2\right|=\sqrt{5}-2\)

c.\(\sqrt{14+6\sqrt{5}}=\sqrt{\left(\sqrt{5}+3\right)^2}=\left|\sqrt{5}+3\right|=\sqrt{5}+3\)

d.\(\sqrt{17-12\sqrt{2}}=\sqrt{\left(2\sqrt{2}-3\right)^2}=\left|2\sqrt{2}-3\right|=3-2\sqrt{2}\)