\(\dfrac{9\sqrt{5}+3\sqrt{27}}{\sqrt{5}+\sqrt{3}}=\dfrac{9\sqrt{5}+9\sqrt{3}}{\sqrt{5}+\sqrt{3}}=\dfrac{9\left(\sqrt{5}+\sqrt{3}\right)}{\sqrt{5}+\sqrt{3}}=9\)

b. 

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

\(=\sqrt{3-\sqrt{5}}.\sqrt{9-5}+\sqrt{3+\sqrt{5}}.\sqrt{9-5}\)

\(=\sqrt{12-4\sqrt{5}}+\sqrt{12+4\sqrt{5}}\)

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

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

c.

\(\dfrac{a-\sqrt{b}}{\sqrt{b}}:\dfrac{\sqrt{b}}{a+\sqrt{b}}=\dfrac{\left(a-\sqrt{b}\right)\left(a+\sqrt{b}\right)}{\sqrt{b}.\sqrt{b}}=\dfrac{a^2-b}{b}\)

a)A=\(2\sqrt{3}-8\sqrt{3}+7\sqrt{3}=\sqrt{3}\)

b)B\(=\sqrt{\left(3-\sqrt{5}\right)^2}+\sqrt{\left(2-\sqrt{5}\right)^2}=3-\sqrt{5}+\sqrt{5}-2=1\)

d)\(=\dfrac{\left(5+\sqrt{5}\right)\left(\sqrt{5}-2\right)}{1}+1-\sqrt{5}-\dfrac{11\left(2\sqrt{5}-3\right)}{11}=5\sqrt{5}+5-10-2\sqrt{5}+1-\sqrt{5}-2\sqrt{5}+3=-1\)

\(a,\dfrac{2}{\sqrt{3}-\sqrt{5}}+\dfrac{3-2\sqrt{3}}{\sqrt{3}-2}\)

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

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

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

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

\(b,\dfrac{5-\sqrt{5}}{\sqrt{5}-1}+\dfrac{\sqrt{5}-\sqrt{3}}{\sqrt{5}+3}\)

\(=\dfrac{\sqrt{5}\left(\sqrt{5}-1\right)}{\sqrt{5}-1}+\dfrac{\left(\sqrt{5}-\sqrt{3}\right)^2}{\left(\sqrt{5}-\sqrt{3}\right)\left(\sqrt{5}+\sqrt{3}\right)}\)

\(=\sqrt{5}+\dfrac{5-2\sqrt{15}+3}{5-3}\)

\(=\dfrac{2\sqrt{5}+8-2\sqrt{15}}{2}\)

\(=\dfrac{2\cdot\left(\sqrt{5}+4-\sqrt{15}\right)}{2}\)

\(=\sqrt{5}-\sqrt{15}+4\)

#\(Toru\)

a) Ta có: \(A=\sqrt{\sqrt{3}+\sqrt{2}}\cdot\sqrt{\sqrt{3}-\sqrt{2}}\)

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

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

b) Ta có: \(B=\sqrt{5-2\sqrt{6}}+\sqrt{5+2\sqrt{6}}\)

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

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

\(=2\sqrt{3}\)

`A=sqrt{sqrt3+sqrt2}.sqrt{sqrt3-sqrt2}`

`=sqrt{(sqrt3+sqrt2)(sqrt3-sqrt2)}`

`=sqrt{3-2}=1`

`b)B=sqrt{5-2sqrt6}+sqrt{5+2sqrt6}`

`=sqrt{3-2sqrt6+2}+sqrt{3+2sqrt6+2}`

`=sqrt{(sqrt3-sqrt2)^2}+sqrt{(sqrt3+sqrt2)^2}`

`=sqrt3-sqrt2+sqrt3+sqrt2=2sqrt3`

`c)C=3-sqrt{3-sqrt5}`

`=3-sqrt{(6-2sqrt5)/2}`

`=3-sqrt{(sqrt5-1)^2/2}`

`=3-(sqrt5-1)/sqrt2`

`=3-(sqrt{10}-sqrt2)/2`

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

Ta có: \(\frac{\sqrt2\left(3+\sqrt5\right)}{2\sqrt2+\sqrt{3+\sqrt5}}\)

\(=\frac{2\left(3+\sqrt5\right)}{4+\sqrt{6+2\sqrt5}}\)

\(=\frac{6+2\sqrt5}{4+\sqrt{\left(\sqrt5+1\right)^2}}=\frac{\left(\sqrt5+1\right)^2}{4+\sqrt5+1}\)

\(=\frac{\left(\sqrt5+1\right)^2}{5+\sqrt5}=\frac{\left(\sqrt5+1\right)^2}{\sqrt5\left(\sqrt5+1\right)}=\frac{\sqrt5+1}{\sqrt5}\)

Ta có: \(\frac{\sqrt2\left(3-\sqrt5\right)}{2\sqrt2-\sqrt{3-\sqrt5}}\)

\(=\frac{2\left(3-\sqrt5\right)}{4-\sqrt{6-2\sqrt5}}\)

\(=\frac{6-2\sqrt5}{4-\sqrt{\left(\sqrt5-1\right)^2}}=\frac{\left(\sqrt5-1\right)^2}{4-\sqrt5+1}\)

\(=\frac{\left(\sqrt5-1\right)^2}{5-\sqrt5}=\frac{\left(\sqrt5-1\right)^2}{\sqrt5\left(\sqrt5-1\right)}=\frac{\sqrt5-1}{\sqrt5}\)

Ta có: \(A=\frac{\sqrt2\left(3+\sqrt5\right)}{2\sqrt2+\sqrt{3+\sqrt5}}+\frac{\sqrt2\left(3-\sqrt5\right)}{2\sqrt2-\sqrt{3-\sqrt5}}\)

\(=\frac{\sqrt5+1+\sqrt5-1}{\sqrt5}=\frac{2\sqrt5}{\sqrt5}\)

=2

1) \(A=2\sqrt{5}-6\sqrt{2}+3\sqrt{5}=5\sqrt{5}-6\sqrt{2}\)

2) \(B=\dfrac{30\left(\sqrt{7}+1\right)}{7-1}+\dfrac{15\left(\sqrt{7}-2\right)}{7-4}=5\sqrt{7}+5+5\sqrt{7}-10=-5+10\sqrt{7}\)

3) \(C=\left(3-\dfrac{\sqrt{5}\left(\sqrt{5}-1\right)}{\sqrt{5}-1}\right)\left(3+\dfrac{\sqrt{5}\left(\sqrt{5}+1\right)}{\sqrt{5}+1}\right)=\left(3-\sqrt{5}\right)\left(3+\sqrt{5}\right)=9-5=4\)

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

\(Xét-biểu-thức:=>T=\sqrt{4+\sqrt{10+2\sqrt{5}}}+\sqrt{4-\sqrt{10+2\sqrt{5}}}.\\ \)
Bình phương T thì được điều bất ngờ =))))))))))))

a: \(A=\sqrt{27}-2\sqrt{48}+3\sqrt{75}\)

\(=3\sqrt3-2\cdot4\sqrt3+3\cdot5\sqrt3\)

\(=3\sqrt3-8\sqrt3+15\sqrt3=18\sqrt3-8\sqrt3=10\sqrt3\)

b: \(B=\sqrt{\left(\sqrt5-2\right)^2}-\sqrt{\left(\sqrt5-3\right)^2}\)

\(=\sqrt5-2-\left|3-\sqrt5\right|\)

\(=\sqrt5-2-3+\sqrt5=-5+2\sqrt5\)

c: \(C=\sqrt{\left(2\sqrt3+1\right)^2}+\sqrt{\left(2\sqrt3-5\right)^2}\)

\(=2\sqrt3+1+\left|5-2\sqrt3\right|\)

\(=2\sqrt3+1+5-2\sqrt3\)

=6

d: \(D=\sqrt{9-4\sqrt5}-\sqrt{14+6\sqrt5}\)

\(=\sqrt{\left(\sqrt5-2\right)^2}-\sqrt{\left(3+\sqrt5\right)^2}\)

\(=\sqrt5-2-\left(3+\sqrt5\right)=\sqrt5-2-3-\sqrt5=-5\)

e: \(E=\frac{4}{\sqrt5-2}-\frac{32}{\sqrt5+1}\)

\(=\frac{4\left(\sqrt5+2\right)}{\left(\sqrt5-2\right)\left(\sqrt5+2\right)}-\frac{32\left(\sqrt5-1\right)}{\left(\sqrt5+1\right)\left(\sqrt5-1\right)}\)

\(=4\left(\sqrt5+2\right)-8\left(\sqrt5-1\right)=4\sqrt5+8-8\sqrt5+8=-4\sqrt5+16\)

f: \(M=\frac{10}{3\sqrt2-4}+\frac{28}{3\sqrt2+2}\)

\(=\frac{10\left(3\sqrt2+4\right)}{\left(3\sqrt2+4\right)\left(3\sqrt2-4\right)}+\frac{28\left(3\sqrt2-2\right)}{\left(3\sqrt2-2\right)\left(3\sqrt2+2\right)}\)

\(=\frac{10\left(3\sqrt2+4\right)}{2}+\frac{28\left(3\sqrt2-2\right)}{18-4}=5\left(3\sqrt2+4\right)+2\left(3\sqrt2-2\right)=15\sqrt2+20+6\sqrt2-4=21\sqrt2+16\)