`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`

Trả lời:

\(\sqrt{15-6\sqrt{6}}+\sqrt{35-12\sqrt{6}}\)

\(=\sqrt{9-6\sqrt{6}+6}+\sqrt{27-12\sqrt{6}+8}\)

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

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

Bài làm:

Ta có: \(\sqrt{15-6\sqrt{6}}+\sqrt{35-12\sqrt{6}}\)

\(=\sqrt{9-6\sqrt{6}+6}+\sqrt{36-12\sqrt{6}+6-7}\)

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

\(=3-\sqrt{6}+\sqrt{\left(-1-\sqrt{6}\right)\left(13-\sqrt{6}\right)}\)

Đến đây thì chịu rồi!

1.\(\sqrt{11-4\sqrt{7}}=\sqrt{11-2\cdot2\sqrt{7}}\)=\(\sqrt{7-2\cdot2\cdot\sqrt{7}+4}=\sqrt{\left(\sqrt{7}-2\right)^2}\)=\(\sqrt{7}-2\)

2.\(\left(3-\sqrt{9}\right)\sqrt{11+6\sqrt{6}}=\left(3-3\right)\sqrt{11+6\sqrt{6}}\)=\(0\cdot\sqrt{11+6\sqrt{6}}=0\)

3.\(\sqrt{15-6\sqrt{6}}+\sqrt{35-12\sqrt{6}}=\)\(\sqrt{15-2\cdot3\cdot\sqrt{6}}+\sqrt{35-2\cdot2\cdot3\cdot\sqrt{2}\cdot\sqrt{3}}\)

                                                      =\(\sqrt{9-2\cdot3\cdot\sqrt{6}+6}+\sqrt{35-2\cdot\left(2\sqrt{2}\right)\left(3\sqrt{3}\right)}\)=\(\sqrt{\left(3-\sqrt{6}\right)^2}+\sqrt{27-2\cdot\left(2\sqrt{2}\right)\left(3\sqrt{3}\right)+8}\)

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

\(\sqrt{15-6\sqrt{6}}+\sqrt{35-12\sqrt{6}}=\sqrt{\left(3-\sqrt{6}\right)^2}+\sqrt{\left(3\sqrt{3}-2\sqrt{2}\right)^2}\)

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

\(\sqrt{17-3\sqrt{32}}+\sqrt{17+3\sqrt{32}}=\sqrt{\left(3-2\sqrt{2}\right)^2}+\sqrt{\left(3+2\sqrt{2}\right)^2}\)

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

\(\sqrt{49-5\sqrt{96}}+\sqrt{49+5\sqrt{96}}=\sqrt{\left(5-2\sqrt{6}\right)^2}+\sqrt{\left(5+2\sqrt{6}\right)^2}\)

\(=5-2\sqrt{6}+5+2\sqrt{6}=10\)

\(\sqrt{13-\sqrt{160}}+\sqrt{53+4\sqrt{90}}=\sqrt{\left(2\sqrt{2}-\sqrt{5}\right)^2}+\sqrt{\left(3\sqrt{5}+2\sqrt{2}\right)^2}\)

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

a: \(\sqrt{15-6\sqrt{6}}+\sqrt{35-12\sqrt{6}}\)

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

b: \(\sqrt{17-3\sqrt{32}}+\sqrt{17+3\sqrt{32}}\)

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

=6

c: Ta có: \(\sqrt{49-5\sqrt{96}}+\sqrt{49+5\sqrt{96}}\)

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

=10

d: Ta có: \(\sqrt{13-\sqrt{160}}+\sqrt{53+4\sqrt{90}}\)

\(=\sqrt{13-4\sqrt{10}}+\sqrt{53+4\sqrt{90}}\)

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

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

\(\dfrac{\sqrt{10}-\sqrt{15}}{\sqrt{8}-\sqrt{12}}\)

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

\(=\dfrac{\sqrt{5}}{\sqrt{4}}\)

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

\(\dfrac{\sqrt{6}-\sqrt{15}}{\sqrt{35}-\sqrt{14}}\)

\(=\dfrac{\sqrt{3}\left(\sqrt{2}-\sqrt{5}\right)}{\sqrt{7}\left(\sqrt{5}-\sqrt{2}\right)}\)

\(=-\dfrac{\sqrt{3}\left(\sqrt{5}-\sqrt{2}\right)}{\sqrt{7}\left(\sqrt{5}-\sqrt{2}\right)}\)

\(=-\dfrac{\sqrt{3}}{\sqrt{7}}\)

\(=-\dfrac{\sqrt{21}}{7}\)

____________

\(\dfrac{5+\sqrt{5}}{\sqrt{10}+\sqrt{2}}\)

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

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

\(=\dfrac{\sqrt{10}}{2}\)

1)
\(=\sqrt{\left(\sqrt{11}\right)^2-2.\sqrt{11}.\sqrt{3}+\left(\sqrt{3}\right)^2}\)
\(=\sqrt{\left(\sqrt{11}-\sqrt{3}\right)^2}=\sqrt{11}-\sqrt{3}\)
2)
\(=\sqrt{\left(\sqrt{7}\right)^2-2.\sqrt{7}\sqrt{5}+\left(\sqrt{5}\right)^2}=\sqrt{\left(\sqrt{7}-\sqrt{5}\right)^2}=\sqrt{7}-\sqrt{5}\)
3)
\(=\sqrt{\left(\sqrt{11}\right)^2-2.\sqrt{11}\sqrt{5}+\left(\sqrt{5}\right)^2}=\sqrt{\left(\sqrt{11}-\sqrt{5}\right)}=\sqrt{11}-\sqrt{5}\)
4)
\(=\sqrt{3^2-2.3.\sqrt{5}+\left(\sqrt{5}\right)^2}=\sqrt{\left(3-\sqrt{5}\right)^2}=3-\sqrt{5}\)
5)
\(=\sqrt{3^2-2.3.2\sqrt{2}+\left(2\sqrt{2}\right)^2}=\sqrt{\left(3-2\sqrt{2}\right)^2}=3-2\sqrt{2}\)

 

\(\dfrac{\sqrt{15}-\sqrt{6}}{\sqrt{35}-\sqrt{14}}=\dfrac{\sqrt{3}\left(\sqrt{5}-\sqrt{2}\right)}{\sqrt{7}\left(\sqrt{5}-\sqrt{2}\right)}=\sqrt{\dfrac{3}{7}}\)

\(\dfrac{\sqrt{15}-\sqrt{5}}{\sqrt{3}-1}=\dfrac{\sqrt{5}\left(\sqrt{3}-1\right)}{\sqrt{3}-1}=\sqrt{5}\)

\(\dfrac{2\sqrt{8}-\sqrt{12}}{\sqrt{18}-\sqrt{48}}=\dfrac{2\left(\sqrt{8}-\sqrt{3}\right)}{\sqrt{6}\left(\sqrt{3}-\sqrt{8}\right)}=-\dfrac{2\sqrt{6}}{6}\)

`(sqrt 15 - sqrt 6)/(sqrt 35 - sqrt 14)`

`= (sqrt 3 . (sqrt 5 - sqrt 2))/(sqrt 7. (sqrt 5 - sqrt 2))`

`= sqrt3/sqrt 7`

`@ (sqrt 15 - sqrt 5)/(sqrt 3 - 1)`

`= (sqrt 5(sqrt 3 - 1))/(sqrt 3 - 1)`

`= sqrt5`

`@ (2 sqrt 8 - sqrt 12)/(sqrt18 - sqrt 48)`

`= (2(sqrt 8 - sqrt 3)/(sqrt 6(sqrt 3 - sqrt 8))`

`= (-2)/(sqrt 6) = (-2 sqrt 6)/6`