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

b: \(\sqrt{12+2\sqrt{35}}-\sqrt{12-2\sqrt{35}}=\sqrt{7}+\sqrt{5}-\sqrt{7}+\sqrt{5}=2\sqrt{5}\)

c: \(\sqrt{16+6\sqrt{7}}=4+\sqrt{7}\)

d: \(\sqrt{31-12\sqrt{3}}=3\sqrt{3}-2\)

e: \(\sqrt{27+10\sqrt{2}}=5+\sqrt{2}\)

f: \(\sqrt{14+6\sqrt{5}}=3+\sqrt{5}\)

a) Ta có: \(\sqrt{14-2\sqrt{33}}\)

\(=\sqrt{11-2\cdot\sqrt{11}\cdot\sqrt{3}+3}\)

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

\(=\left|\sqrt{11}-\sqrt{3}\right|\)

\(=\sqrt{11}-\sqrt{3}\)(Vì \(\sqrt{11}>\sqrt{3}\))

b) Ta có: \(\sqrt{12-2\sqrt{35}}\)

\(=\sqrt{7-2\cdot\sqrt{7}\cdot\sqrt{5}+5}\)

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

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

\(=\sqrt{7}-\sqrt{5}\)(Vì \(\sqrt{7}>\sqrt{5}\))

c) Ta có: \(\sqrt{16-2\sqrt{55}}\)

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

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

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

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

d) Ta có: \(\sqrt{14-6\sqrt{5}}\)

\(=\sqrt{9-2\cdot3\cdot\sqrt{5}+5}\)

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

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

\(=3-\sqrt{5}\)(Vì \(3>\sqrt{5}\))

e) Ta có: \(\sqrt{17-12\sqrt{2}}\)

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

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

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

\(=3-2\sqrt{2}\)(Vì \(3>2\sqrt{2}\))

a: \(=\left(-\sqrt{5}-\sqrt{7}\right)\cdot\left(\sqrt{7}-\sqrt{5}\right)\)

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

=-2

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

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

\(=\dfrac{\sqrt{3}-1+\sqrt{3}+1}{\sqrt{2}}=\sqrt{6}\)

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

\(=\sqrt{10}-\sqrt{10}+3\sqrt{7}=3\sqrt{7}\)

Câu e mình chịu, bạn 😔😔

f, \(\sqrt{\sqrt{5}+\sqrt{3-\sqrt{29-12\sqrt{5}}}}=\sqrt{\sqrt{5}+\sqrt{3-\sqrt{\left(2\sqrt{5}-3\right)^2}}}=\sqrt{\sqrt{5}+\sqrt{3-2\sqrt{5}+3}}=\sqrt{\sqrt{5}+\sqrt{6-2\sqrt{5}}}=\sqrt{\sqrt{5}+\sqrt{\left(\sqrt{5}-1\right)^2}}=\sqrt{\sqrt{5}+\sqrt{5}-1}=\sqrt{2\sqrt{5}-1}\)

mik sửa lại câu f , tí nhé :

f , \(\sqrt{\sqrt{5}+\sqrt{3-\sqrt{29-12\sqrt{5}}}}\)

a,\(=\sqrt{9-2.3.2\sqrt{2}+8}-\sqrt{9+2.3.2\sqrt{2}+8}\)

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

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

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

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

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

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

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

Bạn xem lại câu 5 xem có sai đề không chứ mình tính mãi không ra

1. \(\sqrt{14+2\sqrt{33}}=\sqrt{\left(\sqrt{11}+\sqrt{3}\right)^2}=\sqrt{11}+\sqrt{3}\)

2. \(\sqrt{29-12\sqrt{5}}=\sqrt{\left(\sqrt{20}-\sqrt{9}\right)^2}=2\sqrt{5}-3=-3+2\sqrt{5}\)

3. \(\sqrt{16+2\sqrt{55}}=\sqrt{\left(\sqrt{11}+\sqrt{5}\right)^2}=\sqrt{11}+\sqrt{5}\)

4. \(\sqrt{13+4\sqrt{10}}=\sqrt{\left(\sqrt{8}+\sqrt{5}\right)^2}=2\sqrt{2}+\sqrt{5}\)

5. \(\sqrt{36+12\sqrt{5}}=\sqrt{\left(\sqrt{30}+\sqrt{6}\right)^2}=\sqrt{30}+\sqrt{6}\)

6. \(\sqrt{21-6\sqrt{6}}=\sqrt{\left(\sqrt{18}-\sqrt{3}\right)^2}=3\sqrt{2}-\sqrt{3}=-\sqrt{3}+3\sqrt{2}\)

P/s: Đây là dạng toán căn lồng căn, được dùng máy tính để biến đổi thành căn bình phương, nếu bạn chưa biết thì search gg nhé.

`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`