Mai mình làm cho nha, giờ đi soạn văn đã...
\(1.\dfrac{\sqrt{3}-1+\sqrt{3}+1}{\left(\sqrt{3}-1\right)\left(\sqrt{3}+1\right)}=\dfrac{2\sqrt{3}}{2}=\sqrt{3}\)
\(2.\dfrac{\sqrt{5}\left(\sqrt{5}+1\right)}{\sqrt{5}\left(\sqrt{5}-1\right)}+\dfrac{\sqrt{5}\left(\sqrt{5}-1\right)}{\sqrt{5}\left(\sqrt{5}+1\right)}=\dfrac{\sqrt{5}+1}{\sqrt{5}-1
}+\dfrac{\sqrt{5}-1}{\sqrt{5}+1}=\dfrac{\sqrt{5}+1+\sqrt{5}-1}{\left(\sqrt{5}+1\right)\left(\sqrt{5}-1\right)}=\dfrac{2\sqrt{5}}{4}=\dfrac{\sqrt{5}}{2}\)
\(3.\left(\dfrac{\sqrt{3}\left(1-\sqrt{2}\right)}{1-\sqrt{2}}+\dfrac{\sqrt{5}\left(\sqrt{3}-1\right)}{1-\sqrt{3}}\right):\dfrac{\sqrt{5}-\sqrt{3}}{2}=\left(\sqrt{3}-\sqrt{5}\right)\cdot\dfrac{2}{\sqrt{5}-\sqrt{3}}=-2\)
\(4.\dfrac{\sqrt{3}\left(\sqrt{2}-1\right)}{1-\sqrt{2}}+\dfrac{\sqrt{3}\left(\sqrt{3}+6\right)}{\sqrt{3}}-\dfrac{13\cdot\left(4-\sqrt{3}\right)}{\left(4+\sqrt{3}\right)\cdot\left(4-\sqrt{3}\right)}=-\sqrt{3}+\sqrt{3}+6-4+\sqrt{3}=\sqrt{3}+2\)
\(5.\dfrac{\sqrt{21}\cdot\left(\sqrt{7}-\sqrt{3}\right)}{\sqrt{7}-\sqrt{3}}+\dfrac{4\cdot\left(5+\sqrt{21}\right)}{\left(5-\sqrt{21}\right)\left(5+\sqrt{21}\right)}=\sqrt{21}+5+\sqrt{21}=2\sqrt{21}+5\)
\(6.\dfrac{\sqrt{3}\left(\sqrt{\sqrt{3}+1}+1\right)}{\left(\sqrt{\sqrt{3}+1}+1\right)\left(\sqrt{\sqrt{3}+1}-1\right)}-\dfrac{\sqrt{3}\left(\sqrt{\sqrt{3}+1}-1\right)}{\left(\sqrt{\sqrt{3}+1}-1\right)\left(\sqrt{\sqrt{3}+1}+1\right)}=\sqrt{\sqrt{3}+1}+1-\sqrt{\sqrt{3}+1}+1=2\)
