\(1,\)

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

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

\(2,\)

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

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

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

\(=-3\)

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

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

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

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

\(=6\)

a: =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,\(\dfrac{1}{7}\)

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

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)(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`

Bài 1:
a.

\(\frac{1}{2\sqrt{2}-3\sqrt{3}}=\frac{2\sqrt{2}+3\sqrt{3}}{(2\sqrt{2}-3\sqrt{3})(2\sqrt{2}+3\sqrt{3})}=\frac{2\sqrt{2}+3\sqrt{3}}{(2\sqrt{2})^2-(3\sqrt{3})^2}=\frac{2\sqrt{2}+3\sqrt{3}}{-19}\)

b.

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

Bài 2.

a. 

\(=\frac{\sqrt{8}(\sqrt{5}+\sqrt{3})}{(\sqrt{5}-\sqrt{3})(\sqrt{5}+\sqrt{3})}=\frac{2\sqrt{2}(\sqrt{5}+\sqrt{3})}{5-3}=\sqrt{2}(\sqrt{5}+\sqrt{3})=\sqrt{10}+\sqrt{6}\)

b.

\(=\sqrt{\frac{(2-\sqrt{3})^2}{(2-\sqrt{3})(2+\sqrt{3})}}=\sqrt{\frac{(2-\sqrt{3})^2}{2^2-3}}=\sqrt{(2-\sqrt{3})^2}=|2-\sqrt{3}|=2-\sqrt{3}\)

Bài 3:

a.

\(M=\left[\frac{15(\sqrt{6}-1)}{(\sqrt{6}+1)(\sqrt{6}-1)}+\frac{4(\sqrt{6}+2)}{(\sqrt{6}-2)(\sqrt{6}+2)}-\frac{12(3+\sqrt{6})}{(3-\sqrt{6})(3+\sqrt{6})}\right](\sqrt{6}+11)\)

\(=\left[\frac{15(\sqrt{6}-1)}{6-1}+\frac{4(\sqrt{6}+2)}{6-2^2}-\frac{12(3+\sqrt{6})}{3^2-6}\right](\sqrt{6}+11)\)

\(=[3(\sqrt{6}-1)+2(\sqrt{6}+2)-4(3+\sqrt{6})](\sqrt{6}+11)=(\sqrt{6}-11)(\sqrt{6}+11)=6-11^2=-115\)

b.

\(N=\left[1-\frac{\sqrt{5}(\sqrt{5}+1)}{\sqrt{5}+1}\right].\left[\frac{\sqrt{5}(\sqrt{5}-1)}{1-\sqrt{5}}-1\right]\)

\(=(1-\sqrt{5})(-\sqrt{5}-1)=(\sqrt{5}-1)(\sqrt{5}+1)=5-1=4\)

\(=a^2+2a+1-a^2+2a-1-3a^2+3\\ =-3a^2+4a+3\)

a) \(\sqrt{\dfrac{3-\sqrt{5}}{3+\sqrt{5}}}\)

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

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

\(=\dfrac{\left|3-\sqrt{5}\right|}{\sqrt{9-5}}\)

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

b) \(\sqrt{\dfrac{2-\sqrt{3}}{2+\sqrt{3}}}\)

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

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

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

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

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

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

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

d: \(=\left(-3+3\sqrt{6}+4+2\sqrt{6}-12-4\sqrt{6}\right)\left(\sqrt{6}+11\right)\)

=(căn 6-11)(căn 6+11)

=6-121=-115

Sao ko bảo chị giải đi, kkkk

Haizzzzz

Câu a)

Câu b) Đề bài sai.

`a)((sqrt(14)-sqrt7)/(1-sqrt2)+(sqrt{15}-sqrt5)/(1-sqrt3)):1/(sqrt7-sqrt5)`

`=((sqrt7(sqrt2-1))/(1-sqrt2)+(sqrt5(sqrt3-1))/(1-sqrt3)).(sqrt7-sqrt5)`

`=(-sqrt7-sqrt5)*(sqrt7-sqrt5)`

`=-(sqrt7+sqrt5)(sqrt7+sqrt5)`

`=-(7-5)=-2`

`b)sqrt2+1/sqrt{5+2sqrt6}+2/sqrt{8+2sqrt{15}}`

`=sqrt2+1/sqrt{3+2sqrt{3}.sqrt2+2}+2/sqrt{5+2sqrt{5}.sqrt3+3}`

`=sqrt2+1/sqrt{(sqrt3+sqrt2)^2}+2/sqrt{(sqrt5+sqrt3)^2}`

`=sqrt2+1/(sqrt3+sqrt2)+2/(sqrt5+sqrt3)`

`=sqrt2+((sqrt3+sqrt2)(sqrt3-sqrt2))/(sqrt3+sqrt2)+((sqrt5+sqrt3)(sqrt5-sqrt3))/(sqrt5+sqrt3)`

`=sqrt2+sqrt3-sqrt2+sqrt5-sqrt3=sqrt5`

a) Ta có: \(\left(\dfrac{\sqrt{14}-\sqrt{7}}{1-\sqrt{2}}+\dfrac{\sqrt{15}-\sqrt{5}}{1-\sqrt{3}}\right):\dfrac{1}{\sqrt{7}-\sqrt{5}}\)

\(=\left(-\dfrac{\sqrt{7}\left(1-\sqrt{2}\right)}{1-\sqrt{2}}-\dfrac{\sqrt{5}\left(1-\sqrt{3}\right)}{1-\sqrt{3}}\right):\dfrac{1}{\sqrt{7}-\sqrt{5}}\)

\(=-2\)

b) Ta có: \(\sqrt{2}+\dfrac{1}{\sqrt{5+2\sqrt{6}}}+\dfrac{2}{\sqrt{8+2\sqrt{15}}}\)

\(=\sqrt{2}+\dfrac{1}{\sqrt{3}+\sqrt{2}}+\dfrac{2}{\sqrt{5}+\sqrt{3}}\)

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

\(=\sqrt{5}\)

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

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

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

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

=>\(B\simeq0,35\)