Giải thích các bước giải:

$a)\frac{5+\sqrt5}{5-\sqrt5}+\frac{5-\sqrt5}{5+\sqrt5}$

$=\frac{\sqrt5(\sqrt5+1)}{\sqrt5(\sqrt5-1)}+\frac{\sqrt5(\sqrt5-1)}{\sqrt5(\sqrt5+1)}$

$=\frac{(\sqrt5+1)^2}{(\sqrt5-1)(\sqrt5+1)}+\frac{(\sqrt5-1)^2}{(\sqrt5-1)(\sqrt5+1)}$

$=\frac{6+2\sqrt5+6-2\sqrt5}{(\sqrt5-1)(\sqrt5+1)}$

$=\frac{12}{5-1}$

$=\frac{12}{4}=3$

$b)\frac{3}{\sqrt5+\sqrt2}+\frac{1}{\sqrt2-1}-\frac{4}{3-\sqrt5}$

$=\frac{3(\sqrt2-1)(3-\sqrt5)}{(\sqrt5+\sqrt2)(\sqrt2-1)(3-\sqrt5)}+\frac{(\sqrt5+\sqrt2)(3-\sqrt5)}{(\sqrt5+\sqrt2)(\sqrt2-1)(3-\sqrt5)}-\frac{4(\sqrt5+\sqrt2)(\sqrt2-1)}{(\sqrt5+\sqrt2)(\sqrt2-1)(3-\sqrt5)}$

$=\frac{9\sqrt2-3\sqrt{10}+3\sqrt5-9}{(\sqrt5+\sqrt2)(\sqrt2-1)(3-\sqrt5)}+\frac{3\sqrt5-5+3\sqrt2-\sqrt{10}}{(\sqrt5+\sqrt2)(\sqrt2-1)(3-\sqrt5)}-\frac{4\sqrt{10}-4\sqrt5+8-4\sqrt2}{(\sqrt5+\sqrt2)(\sqrt2-1)(3-\sqrt5)}$

$=\frac{9\sqrt2-3\sqrt{10}+3\sqrt5-9+3\sqrt5-5+3\sqrt2-\sqrt{10}-(4\sqrt{10}-4\sqrt5+8-4\sqrt2)}{(\sqrt5+\sqrt2)(\sqrt2-1)(3-\sqrt5)}$

$=\frac{9\sqrt2-3\sqrt{10}+3\sqrt5-9+3\sqrt5-5+3\sqrt2-\sqrt{10}-4\sqrt{10}+4\sqrt5-8+4\sqrt2}{(\sqrt5+\sqrt2)(\sqrt2-1)(3-\sqrt5)}$

$=\frac{16\sqrt2-8\sqrt{10}+10\sqrt5-22}{(\sqrt5+\sqrt2)(\sqrt2-1)(3-\sqrt5)}$

Kho'k tiếng chos ;-; ;-; ;-;

$c)\frac{\sqrt5-2}{5+2\sqrt5}-\frac{1}{2+\sqrt5}+\frac1{\sqrt5}$

$=\frac{\sqrt5-2}{\sqrt5(\sqrt5+2)}-\frac{\sqrt5}{\sqrt5(\sqrt5+2)}+\frac{\sqrt5+2}{\sqrt5(\sqrt5+2)}$

$=\frac{\sqrt5-2-\sqrt5+\sqrt5+2}{\sqrt5(\sqrt5+2)}$

$=\frac{\sqrt5}{\sqrt5(\sqrt5+2)}$

$=\frac{1}{\sqrt5+2}$