a: \(\dfrac{\sqrt{2}+\sqrt{3}+\sqrt{4}}{\sqrt{2}+\sqrt{3}+\sqrt{6}+\sqrt{8}+4}\)

\(=\dfrac{\sqrt{2}+\sqrt{3}+2}{\sqrt{2}+\sqrt{3}+2+2+\sqrt{6}+\sqrt{8}}\)

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

Lời giải:
a.

\(=\frac{\sqrt{5}+2}{(\sqrt{5}-2)(\sqrt{5}+2)}+\frac{4(\sqrt{5}-1)}{(\sqrt{5}-1)(\sqrt{5}+1)}=\frac{\sqrt{5}+2}{5-2^2}+\frac{4(\sqrt{5}-1)}{5-1}\)

$=\sqrt{5}+2+(\sqrt{5}-1)=2\sqrt{5}+1$
b.

$=\frac{4(\sqrt{3}+1)}{(\sqrt{3}-1)(\sqrt{3}+1)}+\frac{7(3+\sqrt{2})}{(3-\sqrt{2})(3+\sqrt{2})}-2\sqrt{3}$

$=\frac{4(\sqrt{3}+1)}{2}+\frac{7(3+\sqrt{2})}{1}-2\sqrt{3}$
$=2(\sqrt{3}+1)+7(3+\sqrt{2})-2\sqrt{3}$
$=23+7\sqrt{2}$
c.

$=(\frac{4(3+\sqrt{5})}{(3-\sqrt{5})(3+\sqrt{5})}-\frac{\sqrt{5}+2}{(\sqrt{5}-2)(\sqrt{5}+2)}).\frac{7(3+\sqrt{2})}{(3-\sqrt{2})(3+\sqrt{2})}$

$=[(3+\sqrt{5})-(\sqrt{5}+2)].(3+\sqrt{2})$

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

b: Ta có: \(\dfrac{1}{2+\sqrt{3}}+\dfrac{\sqrt{2}}{\sqrt{6}}-\dfrac{2}{3+\sqrt{3}}\)

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

\(=\dfrac{3-\sqrt{3}}{3}\)

\(a,=\dfrac{\left(\sqrt{5}-2\sqrt{3}\right)\left(\sqrt{5}-\sqrt{3}\right)-\left(2\sqrt{5}+\sqrt{3}\right)\left(\sqrt{5}+\sqrt{3}\right)}{\left(\sqrt{5}-\sqrt{3}\right)\left(\sqrt{5}+\sqrt{3}\right)}\\ =\dfrac{11-3\sqrt{15}-13-3\sqrt{15}}{2}=\dfrac{-2-6\sqrt{15}}{2}=-1-3\sqrt{15}\)

\(b,=x\sqrt{2\left(x+1\right)}+\sqrt{\dfrac{2\left(x+1\right)^2}{x+1}}-\sqrt{\dfrac{16\left(x+1\right)}{2}}\\ =x\sqrt{2\left(x+1\right)}+\sqrt{2\left(x+1\right)}-2\sqrt{2\left(x+1\right)}\\ =\sqrt{2\left(x+1\right)}\left(x+1-2\right)=\left(x-1\right)\sqrt{2\left(x+1\right)}\)

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

\(=\dfrac{5-\sqrt{15}-2\sqrt{15}+6}{5-3}-\dfrac{10+2\sqrt{15}+\sqrt{15}+3}{5-3}\)

=\(\dfrac{11-3\sqrt{15}-13-3\sqrt{15}}{2}=\dfrac{-2-6\sqrt{15}}{2}\)

=\(-1-3\sqrt{15}\)

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

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

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

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

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/

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

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

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

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

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

b) Ta có: \(\sqrt{\dfrac{3+\sqrt{5}}{3-\sqrt{5}}}+\sqrt{\dfrac{3-\sqrt{5}}{3+\sqrt{5}}}\)

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

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

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

g: \(=\left(-\sqrt{5}-2\right)\left(\sqrt{5}-2\right)\)

=-(căn 5+2)(căn 5-2)

=-(5-4)=-1

h: \(=\left(\dfrac{4}{3}\sqrt{3}+\sqrt{2}+\dfrac{\sqrt{30}}{3}\right)\left(\dfrac{\sqrt{30}}{5}+\sqrt{2}-\dfrac{4}{5}\sqrt{5}\right)\)

=4/5*căn 10+4/3*căn 6-16/15*căn 15+2/5*căn 15+2-4/5*căn 10+30/15+2/3*căn 15-4/3*căn 6

=4

giúp mình với ạ 

b: Ta có: \(\dfrac{4}{\sqrt{3}+1}+\dfrac{2}{\sqrt{3}-1}-\dfrac{6}{3-\sqrt{3}}\)

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

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

`(5sqrt{1/5}+1/2sqrt{20}-5/4sqrt{4/5}+sqrt{5}):2/5

`=(sqrt5+1/2*2sqrt5-sqrt{5/4}+sqrt5):2/5`

`=(sqrt5+sqrt5+sqrt5-sqrt5/2):2/5`

`=(5/2*sqrt5):2/5`

`=25/4sqrt5`

`1/3sqrt{48}+3sqrt{75}-sqrt{27}-10sqrt{1 1/3}`

`=1/3*4sqrt3+3*5sqrt3-3sqrt3-10sqrt{4/3}`

`=4/sqrt3+15sqrt3-3sqrt3-20/sqrt3`

`=12sqrt3-16/sqrt3`