a: =8/4-1/4=7/4

b: =3/9=1/3

c: =5/7-4/7=1/7

a,\(\dfrac{7}{4}\)

b,\(\dfrac{1}{3}\)

c,\(\dfrac{1}{7}\)

`a)sqrt{8-2sqrt7}+sqrt{16-6sqrt7}`

`=sqrt{(sqrt7-1)^2}+sqrt{(3-sqrt7)^2}`

`=sqrt7-1+3-sqrt7=2`

`b)sqrt{(sqrt7-1)^2}-sqrt{11+4sqrt7}`

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

`=sqrt7-1-2-sqrt7=-3`

a, \(=\sqrt{7-2\sqrt{7}+1}+\sqrt{7-2.3\sqrt{7}+9}\)

\(=\sqrt{\left(\sqrt{7}-1\right)^2}+\sqrt{\left(3-\sqrt{7}\right)^2}=\left|\sqrt{7}-1\right|+\left|3-\sqrt{7}\right|\)

\(=\sqrt{7}-1+3-\sqrt{7}=2\)

\(b,=\left|\sqrt{7}-1\right|-\sqrt{7+2.2\sqrt{7}+4}\)

\(=\left|\sqrt{7}-1\right|-\sqrt{\left(\sqrt{7}+2\right)^2}=\left|\sqrt{7}-1\right|-\left|\sqrt{7}+2\right|\)

\(=\sqrt{7}-1-\sqrt{7}-2=-3\)

a \(\Rightarrow\sqrt{7-2\sqrt{7}+1}+\sqrt{9-6\sqrt{7}+7}=\sqrt{\left(\sqrt{7}-1\right)^2}+\sqrt{\left(3-\sqrt{7}\right)^2}=\sqrt{7}-1+3-\sqrt{7}=2\) b \(\Rightarrow\sqrt{7}-1-\sqrt{7-4\sqrt{7}+4}=\sqrt{7}-1-\sqrt{\left(\sqrt{7}-2\right)^2}=\sqrt{7}-1-\sqrt{7}+2=1\)

`a)A=(3-sqrt5)sqrt{3+sqrt5}+(3+sqrt5)sqrt{3-sqrt5}`

`=sqrt{3-sqrt5}sqrt{3+sqrt5}(sqrt{3+sqrt5}+sqrt{3-sqrt5})`

`=sqrt{9-5}(sqrt{3+sqrt5}+sqrt{3-sqrt5})`

`=2(sqrt{3+sqrt5}+sqrt{3-sqrt5})`

`=sqrt2(sqrt{6+2sqrt5}+sqrt{6-2sqrt5})`

`=sqrt2(sqrt{(sqrt5+1)^2}+sqrt{(sqrt5+1)^2})`

`=sqrt2(sqrt5+1+sqrt5-1)`

`=sqrt{2}.2sqrt5`

`=2sqrt{10}`

`b)B=(5+sqrt{21})(sqrt{14}-sqrt6)sqrt{5-sqrt{21}}`

`=sqrt{5+sqrt{21}}sqrt{5-sqrt{21}}sqrt{5+sqrt{21}}(sqrt{14}-sqrt6)`

`=sqrt{25-21}sqrt{5+sqrt{21}}(sqrt{14}-sqrt6)`

`=2sqrt{5+sqrt{21}}(sqrt{14}-sqrt6)`

`=2sqrt2sqrt{5+sqrt{21}}(sqrt{7}-sqrt3)`

`=2sqrt{10+2sqrt{21}}(sqrt{7}-sqrt3)`

`=2sqrt{(sqrt3+sqrt7)^2}(sqrt{7}-sqrt3)`

`=2(sqrt3+sqrt7)(sqrt{7}-sqrt3)`

`=2(7-3)`

`=8`

`c)C=sqrt{4+sqrt7}-sqrt{4-sqrt7}`

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

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

`=(sqrt7+1)/sqrt2-(sqrt7-1)/2`

`=2/sqrt2=sqrt2`

`d)D=\sqrt{2+sqrt3}+sqrt{14-5sqrt3}+sqrt2`

`=>sqrt2D=sqrt{4+2sqrt3}+sqrt{28-10sqrt3}+2`

`=>sqrt2D=sqrt{(sqrt3+1)^2}+sqrt{(5-sqrt3)^2}+2`

`=>sqrt2D=8`

`=>D=4sqrt2`

1:

a: Vì \(\dfrac{-4}{3}=\dfrac{-4\cdot3}{3\cdot3}=\dfrac{-12}{9}=\dfrac{12}{9}\\ \Rightarrow\dfrac{-4}{3}=\dfrac{12}{9}\)

b: Vì : \(-2\cdot3=-6\\ -6\cdot8=-48\)

nên 2 p/s ko bằng nhau 

\(1,\)

\(a,\sqrt{6-2\sqrt{5}}=\sqrt{\sqrt{5^2}-2.\sqrt{5}.1+1}=\sqrt{\left(\sqrt{5}-1\right)^2}=\left|\sqrt{5}-1\right|=\sqrt{5}-1\)

\(b,\sqrt{8+2\sqrt{7}}=\sqrt{\sqrt{7^2}+2.\sqrt{7}.1+1}=\sqrt{\left(\sqrt{7}+1\right)^2}=\left|\sqrt{7}+1\right|=\sqrt{7}+1\)

\(2,\)

\(a,\sqrt{\left(\sqrt{10}-3\right)^2}-\sqrt{10}\)

\(=\left|\sqrt{10}-3\right|-\sqrt{10}\)

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

\(=-3\)

\(b,\sqrt{\left(5+\sqrt{7}\right)^2}-\sqrt{8-2\sqrt{7}}\)

\(=\left|5+\sqrt{7}\right|-\sqrt{\left(\sqrt{7}-1\right)^2}\)

\(=5+\sqrt{7}-\left|\sqrt{7}-1\right|\)

\(=5+\sqrt{7}-\sqrt{7}+1\)

\(=6\)

a; \(\dfrac{9}{27}\) + \(\dfrac{7}{-49}\)

= \(\dfrac{1}{3}\) - \(\dfrac{1}{7}\)

= \(\dfrac{7}{21}\) - \(\dfrac{3}{21}\)

= \(\dfrac{4}{21}\)

b; - \(\dfrac{12}{10}\) + \(\dfrac{-25}{30}\)

   =  - \(\dfrac{6}{5}\) - \(\dfrac{5}{6}\)

   = -\(\dfrac{36}{30}\) - \(\dfrac{25}{30}\)

  = \(\dfrac{-61}{30}\) 

c; \(\dfrac{-20}{35}\) + \(\dfrac{-16}{-24}\)

 =  - \(\dfrac{4}{7}\) + \(\dfrac{2}{3}\)

= - \(\dfrac{12}{21}\) + \(\dfrac{14}{21}\)

=  \(\dfrac{2}{21}\)

d; - \(\dfrac{21}{77}\) + \(\dfrac{10}{-35}\)

 = - \(\dfrac{3}{11}\) - \(\dfrac{2}{7}\)

 = - \(\dfrac{21}{77}\) -  \(\dfrac{22}{77}\)

= - \(\dfrac{43}{77}\)

b: \(=\left(5+\sqrt{21}\right)\left(\sqrt{7}-\sqrt{3}\right)\sqrt{10-2\sqrt{21}}\)

\(=\left(5+\sqrt{21}\right)\left(10-2\sqrt{21}\right)\)

\(=50-10\sqrt{21}+10\sqrt{21}-42=8\)

a: \(A=\sqrt{\sqrt{2}-1}+\sqrt{\sqrt{2}+1}\)

=>\(A^2=\sqrt{2}-1+\sqrt{2}+1+2\sqrt{2-1}=2\sqrt{2}+2\)

=>\(A=\sqrt{2\sqrt{2}+2}\)

Đặt \(B=\sqrt{\sqrt{2}-1}+\sqrt{\sqrt{2}+1}-\sqrt{2+\sqrt{2}}\)

=>\(B=\sqrt{2\sqrt{2}+2}-\sqrt{2+\sqrt{2}}\)

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

=>\(B^2=4+3\sqrt{2}-2\sqrt[4]{2}\left(2+\sqrt{2}\right)\)

=>\(B\simeq0,35\)

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Bài 2: So sánh các phân số sau

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