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

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

\(=\frac{3\left(\sqrt{2}+\sqrt{3}+\sqrt{5}\right)}{2\sqrt{2}}=\frac{3\sqrt{2}\left(\sqrt{2}+\sqrt{3}+\sqrt{5}\right)}{4}\)

\(\frac{2}{\sqrt{3}-5}=\frac{2\left(\sqrt{3}+5\right)}{3-5^2}=\frac{2\left(\sqrt{3}+5\right)}{-22}=\frac{-5-\sqrt{3}}{11}\)

\(\frac{1}{\sqrt{2}+\sqrt{3}+\sqrt{5}}=\frac{\sqrt{2}+\sqrt{3}-\sqrt{5}}{\left(\sqrt{2}+\sqrt{3}+\sqrt{5}\right)\left(\sqrt{2}+\sqrt{3}-\sqrt{5}\right)}\)
\(=\frac{\sqrt{2}+\sqrt{3}-\sqrt{5}}{\left(\sqrt{2}+\sqrt{3}\right)^2-\left(\sqrt{5}\right)^2}=\frac{\sqrt{2}+\sqrt{3}-\sqrt{5}}{2+2\sqrt{2\cdot3}+3-5}=\frac{\sqrt{2}+\sqrt{3}-\sqrt{5}}{5+2\sqrt{6}-5}=\frac{\sqrt{6}\cdot\left(\sqrt{2}+\sqrt{3}-\sqrt{5}\right)}{\sqrt{6}\cdot2\sqrt{6}}=\frac{2\sqrt{3}+3\sqrt{2}-\sqrt{30}}{12}\)

Ta có \(\frac{1}{\sqrt{2}+\sqrt{3}+\sqrt{5}}\) = \(\frac{\sqrt{2}+\sqrt{3}-\sqrt{5}}{5+2\sqrt{6}-5}\)

= \(\frac{\sqrt{6}\left(\sqrt{2}+\sqrt{3}-\sqrt{5}\right)}{12}\)

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

\(A=\frac{1}{\left(\sqrt{5}+2\right)+\sqrt{3}}\)

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

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

\(A=\frac{\sqrt{5}+2-\sqrt{3}}{6-4\sqrt{5}}\)

\(A=\frac{\left(\sqrt{5}+2-\sqrt{3}\right)\left(6+4\sqrt{5}\right)}{\left(6-4\sqrt{5}\right)\left(6+4\sqrt{5}\right)}\)

\(A=\frac{6\sqrt{5}+20+12+8\sqrt{5}-6\sqrt{3}-4\sqrt{15}}{36-16\cdot5}\)

\(A=\frac{14\sqrt{5}+32-6\sqrt{3}-4\sqrt{15}}{-44}\)

\(A=\frac{6\sqrt{3}+4\sqrt{15}-14\sqrt{5}-32}{44}\)

Nhớ k cho mik đó nha ....... rồi kb lun ahihi

có công thức rồi mà cậu,bài này dễ,cậu thử áp dụng công thức xemm..

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

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

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.

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

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