`(2+sqrt2)/(sqrt2+1)=(sqrt2(sqrt2+1))/(sqrt2+1)=sqrt2` $\\$ `(sqrt(15)-sqrt5)/(sqrt3-1)=(sqrt5(sqrt3-1))/(sqrt3-1)=sqrt5` $\\$ `(2sqrt3-sqrt6)/(sqrt8-2)=(2sqrt3-sqrt6)/(2sqrt2-2)=(sqrt6(sqrt2-1))/(2(sqrt2-1))=sqrt6/2` $\\$ `(a-sqrta)/(1-sqrta)=(sqrta(sqrta-1))/(1-sqrta)=-sqrta` $\\$ `(p-2sqrtp)/(sqrtp-2)=(sqrtp(sqrtp-2))/(sqrtp-2)=sqrtp`
\(\dfrac{2+\sqrt{2}}{1+\sqrt{2}}=\sqrt{2}\)
\(\dfrac{2\sqrt{3}-\sqrt{6}}{\sqrt{8}-\sqrt{2}}=\dfrac{\sqrt{6}\left(\sqrt{2}-1\right)}{2\left(\sqrt{2}-1\right)}=\dfrac{\sqrt{6}}{2}\)
a/ \(\dfrac{2+\sqrt{2}}{1+\sqrt{2}}=\dfrac{\sqrt{2}\left(1+\sqrt{2}\right)}{1+\sqrt{2}}=\sqrt{2}\)
b/ \(\dfrac{2\sqrt{3}-\sqrt{6}}{\sqrt{8}-2}=\dfrac{2\sqrt{3}-\sqrt{2.3}}{2\sqrt{2}-2}=\dfrac{\sqrt{2.3}\left(\sqrt{2}-1\right)}{2\left(\sqrt{2}-1\right)}=\dfrac{\sqrt{3}}{\sqrt{2}}=\dfrac{\sqrt{6}}{2}\)
c/ ĐKXĐ : \(a\ge0\)
\(\dfrac{a-\sqrt{a}}{1-\sqrt{a}}=\dfrac{\sqrt{a}\left(\sqrt{a}-1\right)}{1-\sqrt{a}}=-\sqrt{a}\)
d/ \(p\ge0\)
\(\dfrac{p-2\sqrt{p}}{\sqrt{p}-2}=\dfrac{\sqrt{p}\left(\sqrt{p}-2\right)}{\sqrt{p}-2}=\sqrt{p}\)
