a, \(\frac{\sqrt{3}}{\sqrt{\sqrt{3}+1}-1}-\frac{\sqrt{3}}{\sqrt{\sqrt{3}+1}+1}\)
= \(\frac{\sqrt{3}\left(\sqrt{\sqrt{3}+1}+1\right)}{\sqrt{3}+1-1}-\frac{\sqrt{3}\left(\sqrt{\sqrt{3}+1}-1\right)}{\sqrt{3}+1-1}\)
= \(\frac{\sqrt{3}\left(\sqrt{\sqrt{3}+1}+1\right)}{\sqrt{3}}-\frac{\sqrt{3}\left(\sqrt{\sqrt{3}+1}-1\right)}{\sqrt{3}}\)
= \(\sqrt{\sqrt{3}+1}+1-\sqrt{\sqrt{3}+1}+1\)
= 2
b, \(\frac{2\sqrt{3+\sqrt{5-\sqrt{13+\sqrt{48}}}}}{\sqrt{6}+\sqrt{2}}\)
= \(\frac{2\sqrt{3+\sqrt{5-\sqrt{13+2\sqrt{12}}}}}{\sqrt{6}+\sqrt{2}}\)
= \(\frac{2\sqrt{3+\sqrt{5-\sqrt{\left(\sqrt{12}+1\right)^2}}}}{\sqrt{6}+\sqrt{2}}\)
= \(\frac{2\sqrt{3+\sqrt{5-\sqrt{12}-1}}}{\sqrt{6}+\sqrt{2}}\)
= \(\frac{2\sqrt{3+\sqrt{4-\sqrt{12}}}}{\sqrt{6}+\sqrt{2}}\)
= \(\frac{2\sqrt{3+\sqrt{4-2\sqrt{3}}}}{\sqrt{6}+\sqrt{2}}\)
= \(\frac{2\sqrt{3+\sqrt{\left(\sqrt{3}-1\right)^2}}}{\sqrt{6}+\sqrt{2}}\)
= \(\frac{2\sqrt{3+\sqrt{3}-1}}{\sqrt{6}+\sqrt{2}}\) (\(\sqrt{3}>1\))
= \(\frac{2\sqrt{2+\sqrt{3}}}{\sqrt{2}\left(\sqrt{3}+1\right)}\)
\(\frac{\sqrt{2}.\sqrt{2+\sqrt{3}}}{\sqrt{3}+1}\)
= \(\frac{\sqrt{4+2\sqrt{3}}}{\sqrt{3}+1}\)
= \(\frac{\sqrt{\left(\sqrt{3}+1\right)^2}}{\sqrt{3}+1}\)
= \(\frac{\sqrt{3}+1}{\sqrt{3}+1}\)
= 1
Chúc bn học tốt! (Mk làm đầy đủ lắm r nha bn)
