\(\frac{3\sqrt{10}+\sqrt{20}-3\sqrt{6}-\sqrt{12}}{\sqrt{5}-\sqrt{3}}\)

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

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

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

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

\(a)\)\(\sqrt{15-6\sqrt{6}}+\sqrt{33-12\sqrt{6}}\)

\(=\)\(\sqrt{6-6\sqrt{6}+9}+\sqrt{24-12\sqrt{6}+9}\)

\(=\)\(\sqrt{\left(\sqrt{6}+3\right)}+\sqrt{\left(\sqrt{24}+3\right)}\)

\(=\)\(\left|\sqrt{6}+3\right|+\left|\sqrt{24}+3\right|\)

\(=\)\(\sqrt{6}+3+\sqrt{24}+3\)

\(=\)\(\sqrt{6}\left(1+\sqrt{4}\right)+9\)

\(=\)\(3\sqrt{6}+9\)

Chúc bạn học tốt ~ 

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

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

\(=\)\(2-\sqrt{3}+\sqrt{\left(\sqrt{3}-1\right)^2}\) ( vì \(2=\sqrt{4}>\sqrt{3}\) ) 

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

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

\(=\)\(1\)

Chúc bạn học tốt ~ 

PS : mới lớp 8 sai thì thông cảm >.< 

\(\sqrt{\left(\sqrt{9}-\sqrt{6}\right)^2}\) + \(\sqrt{\left(\sqrt{24}-\sqrt{9}\right)^2}\)

\(\sqrt{9}-\sqrt{6}\) + \(\sqrt{24}-\sqrt{9}\) vì  \(\sqrt{9}>\sqrt{6}\),\(\sqrt{24}>\sqrt{9}\)

\(3-\sqrt{6}\) + \(2\sqrt{6}-3\)

= \(\sqrt{6}\)

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

1. Sửa đề:

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

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

2.

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

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

3.

\(\frac{3}{\sqrt{6}-\sqrt{3}}+\frac{4}{\sqrt{7}+\sqrt{3}}=\frac{3(\sqrt{6}+\sqrt{3})}{(\sqrt{6}-\sqrt{3})(\sqrt{6}+\sqrt{3})}+\frac{4(\sqrt{7}-\sqrt{3})}{(\sqrt{7}+\sqrt{3})(\sqrt{7}-\sqrt{3})}\)

\(=\frac{3(\sqrt{6}+\sqrt{3})}{3}+\frac{4(\sqrt{7}-\sqrt{3})}{4}=\sqrt{6}+\sqrt{3}+\sqrt{7}-\sqrt{3}=\sqrt{6}+\sqrt{7}\)

\(\sqrt{28-6\sqrt{3}}-\sqrt{12+6\sqrt{3}}\)

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

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

học tốt ~

\(a,=\left(2\sqrt{6}-4\sqrt{3}\right)\sqrt{6}+12\sqrt{2}=12-12\sqrt{2}+12\sqrt{2}=12\\ b,=\dfrac{6\left(3-\sqrt{3}\right)}{6}+\sqrt{3}=3-\sqrt{3}+\sqrt{3}=3\)