Bài 20:

a) \(\sqrt{9-4\sqrt{5}}\cdot\sqrt{9+4\sqrt{5}}=\sqrt{81-80}=1\)

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

\(=2\left(2-9\right)=2\cdot\left(-7\right)=-14\)

c: \(\sqrt{2}\cdot\sqrt{2-\sqrt{3}}\cdot\left(\sqrt{3}+1\right)\)

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

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

=2

d) \(\sqrt{2-\sqrt{3}}\cdot\left(\sqrt{6}-\sqrt{2}\right)\left(2+\sqrt{3}\right)\)

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

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

\(=8+4\sqrt{3}-4\sqrt{3}-6\)

=2

cảm ơn anh ạ

Bài 2 

b, `\sqrt{3x^2}=x+2`          ĐKXĐ : `x>=0`

`=>(\sqrt{3x^2})^2=(x+2)^2`

`=>3x^2=x^2+4x+4`

`=>3x^2-x^2-4x-4=0`

`=>2x^2-4x-4=0`

`=>x^2-2x-2=0`

`=>(x^2-2x+1)-3=0`

`=>(x-1)^2=3`

`=>(x-1)^2=(\pm \sqrt{3})^2`

`=>` $\left[\begin{matrix} x-1=\sqrt{3}\\ x-1=-\sqrt{3}\end{matrix}\right.$

`=>` $\left[\begin{matrix} x=1+\sqrt{3}\\ x=1-\sqrt{3}\end{matrix}\right.$

Vậy `S={1+\sqrt{3};1-\sqrt{3}}`

mình nghĩ ĐKXĐ là như này : 

x+2≥0

➩ x≥-2

có phải k

\(\text{Theo đề bài: }=\dfrac{3\sqrt{2}+6\sqrt{3}+2\sqrt{5}-\sqrt{6}}{2}\)

c/ = \(\sqrt{13+30\sqrt{2+\sqrt{8+2.2\sqrt{2}+1}}}\)

\(=\sqrt{13+30\sqrt{3+2\sqrt{2}}}\)

\(=\sqrt{43+30\sqrt{2}}\)

\(=\sqrt{25+2.3.5.\sqrt{2}+18}\)

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

d/ \(=\sqrt{13+6\sqrt{4+\sqrt{9-4\sqrt{2}}}}\)

\(=\sqrt{13+6\sqrt{4+2\sqrt{2}-1}}\)

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

\(=\sqrt{19+6\sqrt{2}}\)

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

a) \(\sqrt{3+\sqrt{5}}\left(\sqrt{10}+\sqrt{2}\right)\left(3-\sqrt{5}\right)\)

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

\(=\left(9-5\right).\sqrt{3-\sqrt{5}}.\sqrt{2}\left(\sqrt{5}+1\right)\)

\(=4.\sqrt{6-2\sqrt{5}}.\left(\sqrt{5}+1\right)\)

\(=4.\sqrt{5-2\sqrt{5}+1}.\left(\sqrt{5}+1\right)\)

\(=4.\sqrt{\left(\sqrt{5}-1\right)^2}.\left(\sqrt{5}+1\right)\)

\(=4.\left(\sqrt{5}-1\right)\left(\sqrt{5}+1\right)=4.\left(5-1\right)=16\)

b) \(2\sqrt{4+\sqrt{6-2\sqrt{5}}}.\left(\sqrt{10}-\sqrt{2}\right)\)

\(=2\sqrt{4+\sqrt{5-2\sqrt{5}+1}}.\left(\sqrt{10}-\sqrt{2}\right)\)

\(=2\sqrt{4+\sqrt{\left(\sqrt{5}-1\right)^2}}.\left(\sqrt{10}-\sqrt{2}\right)\)

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

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

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

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

\(=2.\left(5-1\right)=2.4=8\)

\(b,\sqrt{2}.\sqrt{7+3\sqrt{5}}-\dfrac{4}{\sqrt{5}-1}\\ =\sqrt{14+6\sqrt{5}}-\dfrac{4}{\sqrt{5}-1}\\ =\sqrt{\sqrt{5^2}+2.3\sqrt{5}+3^2}-\dfrac{4}{\sqrt{5}-1}\\ =\sqrt{\left(\sqrt{5}+3\right)^2}-\dfrac{4}{\sqrt{5}-1}\\ =\left|\sqrt{5}+3\right|-\dfrac{4}{\sqrt{5}-1}\\ =\dfrac{\left(\sqrt{5}+3\right)\left(\sqrt{5}-1\right)-4}{\sqrt{5}-1}\\ =\dfrac{2+2\sqrt{5}-4}{\sqrt{5}-1}\\ =\dfrac{-2+2\sqrt{5}}{\sqrt{5}-1}\\ =\dfrac{2\left(-1+\sqrt{5}\right)}{\sqrt{5}-1}\\ =2\)

\(c,\sqrt{27}-6\sqrt{\dfrac{1}{3}}+\dfrac{\sqrt{3}-3}{\sqrt{3}}\\ =3\sqrt{3}-\dfrac{6}{\sqrt{3}}+\dfrac{\sqrt{3}-3}{\sqrt{3}}\)

\(=\dfrac{3\sqrt{3}.\sqrt{3}-6+\sqrt{3}-3}{\sqrt{3}}\\ =\dfrac{9-6+\sqrt{3}-3}{\sqrt{3}}\\ =\dfrac{\sqrt{3}}{\sqrt{3}}\\ =1\)

\(d,\dfrac{9-2\sqrt{3}}{3\sqrt{6}-2\sqrt{2}}\\ =\dfrac{\left(9-2\sqrt{3}\right)\left(3\sqrt{6}+2\sqrt{2}\right)}{\left(3\sqrt{6}-2\sqrt{2}\right)\left(3\sqrt{6}+2\sqrt{2}\right)}\\ =\dfrac{27\sqrt{6}+18\sqrt{2}-18\sqrt{2}-4\sqrt{6}}{\left(3\sqrt{6}\right)^2-\left(2\sqrt{2}\right)^2}\\ =\dfrac{23\sqrt{6}}{54-8}\\ =\dfrac{23\sqrt{6}}{46}\\ =\dfrac{\sqrt{6}}{2}\)

Giải chi tiết từng bước nha