\(1,\)

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

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

\(2,\)

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

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

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

\(=-3\)

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

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

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

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

\(=6\)

`a)Đặt \, A=sqrt{21+3sqrt{48}}-sqrt{21-3sqrt{48}}`

Vì `21+3sqrt{48}>21-3sqrt{48}`

`=>sqrt{21+3sqrt{48}}-sqrt{21-3sqrt{48}}>0`

Hay `A>0`

`<=>A^2=21+3sqrt{48}+21-3sqrt{48}-2sqrt{21^2-9.48}`

`<=>A^2=42-2sqrt{9}=32-2.3=26`

`<=>A=sqrt{26}(do \ A>0)`

b)Chắc đề là như này:

`sqrt{7-2sqrt{10}}-sqrt{7+2sqrt{10}}`

`=sqrt{5-2sqrt{5}.sqrt2+2}-sqrt{5+2sqrt{5}.sqrt2+2}`

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

`=sqrt5-sqrt2-sqrt5-sqrt2=-2sqrt2`

a) Ta có: \(\sqrt{21+3\sqrt{48}}-\sqrt{21-3\sqrt{48}}\)

\(=\sqrt{12+2\cdot2\sqrt{3}\cdot3+9}-\sqrt{12-2\cdot2\sqrt{3}\cdot3+9}\)

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

=6

b) Ta có: \(\sqrt{7-2\sqrt{10}}-\sqrt{7+2\sqrt{10}}\)

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

\(=2\sqrt{2}\)

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

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

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

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

b) \(B=\sqrt{11-6\sqrt{2}}-\sqrt{3-2\sqrt{2}}\)

\(=\sqrt{3^2-2.3.\sqrt{2}+\left(\sqrt{2}\right)^2}-\sqrt{\left(\sqrt{2}\right)^2-2.\sqrt{2}.1+1^2}\)

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

\(=3-\sqrt{2}-\sqrt{2}+1=4-2\sqrt{2}\)

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

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

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

\(=\left(\sqrt{5}+\sqrt{3}\right)\left(\sqrt{5}-\sqrt{2}\right)=5-\sqrt{10}+\sqrt{15}-\sqrt{6}\)

`a)(5sqrt2-2sqrt5)/(sqrt5-sqrt2)+6/(2-sqrt{10})`

`=(sqrt{10}(sqrt5-sqrt2))/(sqrt5-sqrt2)+(6(2+sqrt{10}))/(4-10)`

`=sqrt{10}-(2+sqrt{10})`

`=-2`

`b)6/(sqrt5-1)+7/(1-sqrt3)-2/(sqrt3-sqrt5)`

`=(6(sqrt5+1))/(5-1)+(7(1+sqrt3))/(1-3)-(2(sqrt3+sqrt5))/(3-5)`

`=(6(sqrt5+1))/4-(7+7sqrt3)/2+sqrt3+sqrt5`

`=(3sqrt5+3)/2-(7+7sqrt3)/2+sqrt3+sqrt5`

`=(3sqrt5+3-7-7sqrt3+2sqrt3+2sqrt5)/2`

`=(5sqrt5-5sqrt3-4)/2`

i: =-12*căn 3/2căn 3=-6

h: =72căn 2/12căn 2=6

g: =25căn 12/5căn 6=5căn 2

f: =(15:5)*căn 6:3=3căn 2

d: =-1/2*6*căn 10=-3căn 10

\(a,\dfrac{\sqrt{10}-\sqrt{15}}{\sqrt{8}-\sqrt{12}}\\ =\dfrac{\sqrt{2}.\sqrt{5}-\sqrt{3}.\sqrt{5}}{\sqrt{2}.\sqrt{4}-\sqrt{3}.\sqrt{4}}\\ =\dfrac{\sqrt{5}\left(\sqrt{2}-\sqrt{3}\right)}{\sqrt{4}\left(\sqrt{2}-\sqrt{3}\right)}\\ =\dfrac{\sqrt{5}}{\sqrt{2^2}}\\ =\dfrac{\sqrt{5}}{2}\)

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

\(c,\dfrac{5+\sqrt{5}}{\sqrt{10}+\sqrt{2}}\\ =\dfrac{\sqrt{5}\left(\sqrt{5}+1\right)}{\sqrt{2}\left(\sqrt{5}+1\right)}=\dfrac{\sqrt{5}}{\sqrt{2}}\)

\(a,=\dfrac{\sqrt{2}.\sqrt{5}-\sqrt{3}.\sqrt{5}}{2\sqrt{2}-2\sqrt{3}}\\ =\dfrac{\sqrt{5}.\left(\sqrt{2}-\sqrt{3}\right)}{2\left(\sqrt{2}-\sqrt{3}\right)}\\ =\dfrac{\sqrt{5}}{2}\)

\(b,=\dfrac{\sqrt{3}.\sqrt{5}-\sqrt{2}.\sqrt{3}}{\sqrt{5}.\sqrt{7}-\sqrt{2}.\sqrt{7}}\\ =\dfrac{\sqrt{3}\left(\sqrt{5}-\sqrt{2}\right)}{\sqrt{7}\left(\sqrt{5}-\sqrt{2}\right)}\\ =\dfrac{\sqrt{3}}{\sqrt{7}}=\dfrac{\sqrt{21}}{7}\)

\(c,=\dfrac{\sqrt{5}.\sqrt{5}+\sqrt{5}}{\sqrt{2}.\sqrt{5}+\sqrt{2}}\\ =\dfrac{\sqrt{5}\left(\sqrt{5}+1\right)}{\sqrt{2}\left(\sqrt{5}+1\right)}\\ =\dfrac{\sqrt{5}}{\sqrt{2}}=\dfrac{\sqrt{10}}{2}\)

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

a)\(=\left(\sqrt{10}+3\sqrt{2}\right)\sqrt{\left(3-\sqrt{5}\right)^2}=\left(\sqrt{10}+3\sqrt{2}\right)\left(3-\sqrt{5}\right)=3\sqrt{10}-5\sqrt{2}+9\sqrt{2}-3\sqrt{10}=4\sqrt{2}\)

b) \(=\sqrt{\left(3-\sqrt{5}\right)\left(3+\sqrt{5}\right)}=\sqrt{9-5}=\sqrt{4}=2\)

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 \(\Rightarrow\sqrt{7}-1-\sqrt{7-4\sqrt{7}+4}=\sqrt{7}-1-\sqrt{\left(\sqrt{7}-2\right)^2}=\sqrt{7}-1-\sqrt{7}+2=1\)