$A=\frac{\sqrt{5+\sqrt{17}}-\sqrt{5-\sqrt{17}}-\sqrt{10-4\sqrt{2}}+4}{\sqrt{3+\sqrt{5}}-\sqrt{3-\sqrt{5}}+2-\sqrt{2}}$
Ta có : $\,{{\left( \sqrt{5+\sqrt{17}}-\sqrt{5-\sqrt{17}} \right)}^{2}}$
$={{\left( \sqrt{5+\sqrt{17}} \right)}^{2}}-2\sqrt{\left( 5+\sqrt{17} \right)\left( 5-\sqrt{17} \right)}+{{\left( \sqrt{5-\sqrt{17}} \right)}^{2}}$
$=10-2\sqrt{8}$
$=10-4\sqrt{2}$
$={{\left( \sqrt{10-4\sqrt{2}} \right)}^{2}}$
Do đó $\sqrt{5+\sqrt{17}}-\sqrt{5-\sqrt{17}}=\sqrt{10-4\sqrt{2}}$$\Leftrightarrow \sqrt{5+\sqrt{17}}-\sqrt{5-\sqrt{17}}-\sqrt{10-4\sqrt{2}}=0$
Tử số = $\sqrt{5+\sqrt{17}}-\sqrt{5-\sqrt{17}}-\sqrt{10-4\sqrt{2}}+4=4$
$A=\frac{\sqrt{5+\sqrt{17}}-\sqrt{5-\sqrt{17}}-\sqrt{10-4\sqrt{2}}+4}{\sqrt{3+\sqrt{5}}-\sqrt{3-\sqrt{5}}+2-\sqrt{2}}$
$A=\frac{\sqrt{2}.4}{\sqrt{6+2\sqrt{5}}-\sqrt{6-2\sqrt{5}}+2\sqrt{2}-2}$
$A=\frac{\sqrt{2}.4}{\sqrt{{{\left( \sqrt{5}+1 \right)}^{2}}}-\sqrt{{{\left( \sqrt{5}-1 \right)}^{2}}+2\sqrt{2}-2}}$
$A=\frac{4\sqrt{2}}{\sqrt{5}+1-\left( \sqrt{5}-1 \right)+2\sqrt{2}-2}$
$A=\frac{4\sqrt{2}}{2\sqrt{2}}$
$A=2$

$B=\frac{2+\sqrt{3}}{\sqrt{2}+\sqrt{2+\sqrt{3}}}+\frac{2-\sqrt{3}}{\sqrt{2}-\sqrt{2-\sqrt{3}}}$
$B=\frac{2+\sqrt{3}}{\sqrt{2}+\sqrt{\frac{4+2\sqrt{3}}{2}}}+\frac{2-\sqrt{3}}{\sqrt{2}-\sqrt{\frac{4-2\sqrt{3}}{2}}}$
$B=\frac{2+\sqrt{3}}{\sqrt{2}+\sqrt{\frac{{{\left( \sqrt{3}+1 \right)}^{2}}}{2}}}+\frac{2-\sqrt{3}}{\sqrt{2}-\sqrt{\frac{{{\left( \sqrt{3}+1 \right)}^{2}}}{2}}}$
$B=\frac{2+\sqrt{3}}{\sqrt{2}+\frac{\sqrt{3}+1}{\sqrt{2}}}+\frac{2-\sqrt{3}}{\sqrt{2}-\frac{\sqrt{3}-1}{\sqrt{2}}}$
$B=\frac{2+\sqrt{3}}{\frac{3+\sqrt{3}}{\sqrt{2}}}+\frac{2-\sqrt{3}}{\frac{3-\sqrt{3}}{\sqrt{2}}}$
$B=\frac{\sqrt{2}\left( 2+\sqrt{3} \right)}{3+\sqrt{3}}+\frac{\sqrt{2}\left( 2-\sqrt{3} \right)}{3-\sqrt{3}}$
$B=\sqrt{\frac{2}{3}}.\left( \frac{2+\sqrt{3}}{\sqrt{3}+1}+\frac{2-\sqrt{3}}{\sqrt{3}-1} \right)$
$B=\sqrt{\frac{2}{3}}.\left( \frac{(2+\sqrt{3})(\sqrt{3}-1)}{(\sqrt{3}+1)\left( \sqrt{3}-1 \right)}+\frac{(2-\sqrt{3})(\sqrt{3}+1)}{(\sqrt{3}-1)\left( \sqrt{3}+1 \right)} \right)$
$B=\sqrt{\frac{2}{3}}.\left( \frac{\sqrt{3}+1}{3-1}+\frac{\sqrt{3}-1}{3-1} \right)$
$B=\sqrt{\frac{2}{3}}.\frac{2\sqrt{3}}{2}=\sqrt{2}$