\(\dfrac{8}{\sqrt{5}-1}-\dfrac{22}{4+\sqrt{5}}+\dfrac{\sqrt{15}+2\sqrt{5}}{2+\sqrt{3}}\)
\(=\dfrac{8\left(\sqrt{5}+1\right)}{\left(\sqrt{5}-1\right)\left(\sqrt{5}+1\right)}-\dfrac{22\left(4-\sqrt{5}\right)}{\left(\sqrt{5}+4\right)\left(4-\sqrt{5}\right)}+\dfrac{\sqrt{5}\left(\sqrt{3}+2\right)}{2+\sqrt{3}}\)
\(=\dfrac{8\sqrt{5}+8}{5-1}-\dfrac{88-22\sqrt{5}}{16-5}+\sqrt{5}\)
\(=\dfrac{8\sqrt{5}+8}{4}-\dfrac{88-22\sqrt{5}}{11}+\sqrt{5}\)
\(=2\sqrt{5}+2-8+2\sqrt{5}+\sqrt{5}=5\sqrt{5}-6\)

\(M=\dfrac{8\left(\sqrt{5}+\sqrt{3}\right)}{2}-\dfrac{7\left(2+\sqrt{3}\right)}{4-3}+\dfrac{4\left(\sqrt{2}+1\right)}{2-1}+\dfrac{\sqrt{15}\left(\sqrt{3}-1\right)}{\sqrt{15}}\)

\(=4\left(\sqrt{5}+\sqrt{3}\right)-14-7\sqrt{3}+4\sqrt{2}+4+\sqrt{3}-1\)

\(=4\sqrt{5}+4\sqrt{3}-6\sqrt{3}+4\sqrt{2}-11\)

\(=4\sqrt{5}-2\sqrt{3}+4\sqrt{2}-11\)

\(M=\dfrac{8\left(\sqrt{5}+\sqrt{3}\right)}{5-3}+\dfrac{7\left(\sqrt{3}+2\right)}{3-4}+\dfrac{4\left(\sqrt{2}+1\right)}{2-1}+\dfrac{\sqrt{15}\left(\sqrt{3}-1\right)}{\sqrt{15}}\)

\(=4\sqrt{5}+4\sqrt{3}-7\sqrt{3}-14+4\sqrt{2}+4+\sqrt{3}-1\)

\(=4\sqrt{5}-2\sqrt{3}+4\sqrt{2}-11\)

\(a,=\dfrac{3\left(\sqrt{5}+\sqrt{2}\right)}{5-2}+\dfrac{4\left(\sqrt{6}-\sqrt{2}\right)}{6-2}+\dfrac{3.\left(\sqrt{6}-\sqrt{5}\right)}{6-5}\\ =\dfrac{3\left(\sqrt{5}+\sqrt{2}\right)}{3}+\dfrac{4\left(\sqrt{6}-\sqrt{2}\right)}{4}+3\left(\sqrt{6}-\sqrt{5}\right)\\ =\sqrt{5}+\sqrt{2}+\sqrt{6}-\sqrt{2}+3\sqrt{6}-3\sqrt{5}\\ =4\sqrt{6}-2\sqrt{5}\)

\(b,=\dfrac{3\left(\sqrt{5}+\sqrt{2}\right)}{5-2}-\dfrac{1}{\sqrt{5-2\sqrt{6}}}-\dfrac{\sqrt{2}.\sqrt{2}}{\sqrt{2}\sqrt{4+\sqrt{15}}}\\ =\dfrac{3\left(\sqrt{5}+\sqrt{2}\right)}{3}-\dfrac{1}{\sqrt{\left(\sqrt{3}-\sqrt{2}\right)^2}}-\dfrac{2}{\sqrt{8+2.\sqrt{3}.\sqrt{5}}}\\ =\sqrt{5}+\sqrt{2}-\dfrac{1}{\left|\sqrt{3}-\sqrt{2}\right|}-\dfrac{2}{\sqrt{\left(\sqrt{5}+\sqrt{3}\right)^2}}\\ =\sqrt{5}+\sqrt{2}-\dfrac{1}{\sqrt{3}-\sqrt{2}}-\dfrac{2}{\left|\sqrt{5}+\sqrt{3}\right|}\)

\(=\sqrt{5}+\sqrt{2}-\dfrac{\sqrt{3}+\sqrt{2}}{3-2}-\dfrac{2.\left(\sqrt{5}-\sqrt{3}\right)}{5-3}\\ =\sqrt{5}+\sqrt{2}-\sqrt{3}-\sqrt{2}-\dfrac{2.\left(\sqrt{5}-\sqrt{3}\right)}{2}\\ =\sqrt{5}+\sqrt{2}-\sqrt{3}-\sqrt{2}-\sqrt{5}+\sqrt{3}\\ =0\)

a: \(=\dfrac{3\left(\sqrt{5}+\sqrt{2}\right)}{3}+\dfrac{4\left(\sqrt{6}-\sqrt{2}\right)}{4}+\dfrac{3\left(\sqrt{6}-\sqrt{5}\right)}{1}\)

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

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

b: \(=\dfrac{3\left(\sqrt{5}+\sqrt{2}\right)}{3}-\dfrac{1}{\sqrt{5-2\sqrt{6}}}+\dfrac{2}{\sqrt{8+2\sqrt{15}}}\)

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

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

=2căn 5-2căn 3

`c)1/(2sqrt2)-3/2sqrt{4,5}+2/5sqrt{50}`

`=1/(2sqrt2)-3/2sqrt{9/2}+2/5sqrt{25.2}`

`=1/(2sqrt2)-9/(2sqrt2)+2sqrt2`

`=2sqrt2-8/(2sqrt2)`

`=2sqrt2-sqrt2=sqrt2`

`d)4/(3+sqrt5)-8/(1+sqrt5)+15/sqrt5`

`=(4(3-sqrt5))/(9-5)-(8(sqrt5-1))/(5-1)+3sqrt5`

`=3-sqrt5-2(sqrt5-1)+3sqrt5`

`=3+3sqrt5-3sqrt5+2=5`

Lời giải:
1/

\(=\frac{3.\sqrt{5}-\sqrt{5}.\sqrt{15}}{\sqrt{15}-3}=\frac{-\sqrt{5}(\sqrt{15}-3)}{\sqrt{15}-3}=-\sqrt{5}\)

2/

\(=\frac{\sqrt{2+2\sqrt{2.3}+3}}{\sqrt{2}+\sqrt{3}}=\frac{\sqrt{(\sqrt{2}+\sqrt{3})^2}}{\sqrt{2}+\sqrt{3}}=\frac{\sqrt{2}+\sqrt{3}}{\sqrt{2}+\sqrt{3}}=1\)

3/

\(=\frac{2^2+2.2\sqrt{3}+3}{2+\sqrt{3}}=\frac{(2+\sqrt{3})^2}{2+\sqrt{3}}=2+\sqrt{3}\)

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.

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

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

\(A=\left(\dfrac{15-\sqrt{x}}{x-25}+\dfrac{2}{\sqrt{x}+5}\right):\dfrac{\sqrt{x}+1}{\sqrt{x}-5}\)

\(=\dfrac{15-\sqrt{x}+2\sqrt{x}-10}{\left(\sqrt{x}+5\right)\left(\sqrt{x}-5\right)}\cdot\dfrac{\sqrt{x}-5}{\sqrt{x}+1}\)

\(=\dfrac{1}{\sqrt{x}+1}\)

\(A=\left(\dfrac{15-\sqrt{x}}{x-25}+\dfrac{2}{\sqrt{x}+5}\right):\dfrac{\sqrt{x}+1}{\sqrt{x}-5}\left(x\ge0;x\ne25\right)\\ A=\dfrac{15-\sqrt{x}+2\left(\sqrt{x}-5\right)}{\left(\sqrt{x}-5\right)\left(\sqrt{x}+5\right)}\cdot\dfrac{\sqrt{x}-5}{\sqrt{x}+1}\\ A=\dfrac{5+\sqrt{x}}{\sqrt{x}+5}\cdot\dfrac{1}{\sqrt{x}+1}=\dfrac{1}{\sqrt{x}+1}\)

\(C=\left(\dfrac{15-\sqrt{x}}{x-25}+\dfrac{2}{\sqrt{x}+5}\right):\dfrac{\sqrt{x}+1}{\sqrt{x}-5}\left(đk:x\ge0,x\ne25\right)\)

\(=\dfrac{15-\sqrt{x}+2\left(\sqrt{x}-5\right)}{\left(\sqrt{x}-5\right)\left(\sqrt{x}+5\right)}.\dfrac{\sqrt{x}-5}{\sqrt{x}+1}\)

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

\(ĐK:x\ge0;x\ne25\)

\(C=\dfrac{15-\sqrt{x}+2\sqrt{x}-10}{\left(\sqrt{x}-5\right)\left(\sqrt{x}+5\right)}\cdot\dfrac{\sqrt{x}-5}{\sqrt{x}+1}\\ C=\dfrac{\sqrt{x}+5}{\left(\sqrt{x}+5\right)\left(\sqrt{x}+1\right)}=\dfrac{1}{\sqrt{x}+1}\)

DKXD: \(x\ne5;x>0\)

\(C=\left(\dfrac{15-\sqrt[]{x}}{x-25}+\dfrac{2}{\sqrt[]{x}+5}\right):\dfrac{\sqrt[]{x+1}}{\sqrt[]{x}-5}\)
\(C=\left(\dfrac{15-\sqrt[]{x}}{\left(\sqrt[]{x}—5\right)\left(\sqrt{x}+5\right)}+\dfrac{2\left(\sqrt[]{x}-5\right)}{\left(\sqrt[]{x}-5\right)\left(\sqrt{x+5}\right)}\right):\dfrac{\sqrt{x}+1}{\sqrt{x}-5}\)
\(C=\left(\dfrac{15-\sqrt{x}+2\sqrt{x}-10}{\left(\sqrt{x}+5\right)\left(\sqrt{x}-5\right)}\right).\dfrac{\sqrt{x}-5}{\sqrt{x}+1}\)

\(C=\dfrac{5+\sqrt{x}}{\left(\sqrt{x}-5\right)\left(\sqrt{x}+5\right)}.\dfrac{\sqrt{x}-5}{\sqrt{x}+1}\)
\(C=\dfrac{1}{\sqrt{x}+1}\)

\(\sqrt{4+2\sqrt{3}}+\sqrt{4-2\sqrt{3}}+\dfrac{5}{2\sqrt{2}-3}-\dfrac{5}{\sqrt{3}+\sqrt{8}}\)

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

\(=6+2\sqrt{3}\)

\(=\sqrt{3+2\sqrt{2}+1}+\sqrt{3-2\sqrt{2}+1}-\dfrac{5\left(\sqrt{3}+2\sqrt{2}\right)}{\left(\sqrt{3}+2\sqrt{2}\right)\left(\sqrt{3}-2\sqrt{2}\right)}-\dfrac{5\left(\sqrt{3}-2\sqrt{2}\right)}{\left(\sqrt{3}+2\sqrt{2}\right)\left(\sqrt{3}-2\sqrt{2}\right)}\\ =\left|\sqrt[]{3}+1\right|+\left|\sqrt{3}-1\right|-\dfrac{5\left(\sqrt{3}+2\sqrt{2}\right)}{5}-\dfrac{5\left(\sqrt{3}-2\sqrt[]{2}\right)}{5}\\ =\sqrt{3}+1+\sqrt{3}-1-\sqrt{3}-2\sqrt{2}-\sqrt[]{3}+2\sqrt{2}\\ =0\)