7:
a: \(\dfrac{3}{\sqrt{5}}=\dfrac{3\sqrt{5}}{5}\)
b: \(\dfrac{2\sqrt{3}}{\sqrt{2}}=\dfrac{2\sqrt{3}\cdot\sqrt{2}}{2}=\sqrt{6}\)
c: \(\dfrac{2+\sqrt{3}}{2-\sqrt{3}}=\dfrac{\left(2+\sqrt{3}\right)^2}{\left(2-\sqrt{3}\right)\left(2+\sqrt{3}\right)}=\left(2+\sqrt{3}\right)^2=7+4\sqrt{3}\)
d: \(\dfrac{1}{\sqrt{3}+\sqrt{2}}=\dfrac{\sqrt{3}-\sqrt{2}}{3-2}=\sqrt{3}-\sqrt{2}\)
e: \(\dfrac{\sqrt{2}+1}{\sqrt{2}-1}=\dfrac{\left(\sqrt{2}+1\right)^2}{\left(\sqrt{2}-1\right)\left(\sqrt{2}+1\right)}=\left(\sqrt{2}+1\right)^2=3+2\sqrt{2}\)
g: \(\dfrac{3\sqrt{2}}{\sqrt{3}+1}=\dfrac{3\sqrt{2}\left(\sqrt{3}-1\right)}{\left(\sqrt{3}+1\right)\left(\sqrt{3}-1\right)}=\dfrac{3\sqrt{6}-3\sqrt{2}}{2}\)
giúp em làm bài 7 bài 8 câu a và b với ạ thanks
