Question

Bài 1 ạ

Answer 1

\(1,\dfrac{\sqrt{2}}{\sqrt{6}-2}=\dfrac{\sqrt{2}}{\sqrt{2}\left(\sqrt{3}-\sqrt{2}\right)}=\dfrac{1}{\sqrt{3}-\sqrt{2}}=\dfrac{\sqrt{3}+\sqrt{2}}{3-2}=\sqrt{3}+\sqrt{2}\\ ----------\\ 2,\dfrac{6}{\sqrt{12}-\sqrt{3}}=\dfrac{6}{\sqrt{3}\left(\sqrt{4}-1\right)}=\dfrac{6}{\sqrt{3}\left(2-1\right)}=\dfrac{6}{\sqrt{3}}=2\sqrt{3}\\ ----------\\ 3,\dfrac{12}{\sqrt{5}-\sqrt{2}}=\dfrac{12\left(\sqrt{5}+\sqrt{2}\right)}{5-2}=\dfrac{12\left(\sqrt{5}+\sqrt{2}\right)}{3}=4\left(\sqrt{5}+\sqrt{2}\right)=4\sqrt{5}+4\sqrt{2}\\ ----------\)

\(4,\dfrac{\left(\sqrt{2}+2\right)}{\sqrt{2}-2}=\dfrac{\left(\sqrt{2}+2\right)\left(\sqrt{2}+2\right)}{2-4}=\dfrac{\left(\sqrt{2}+2\right)^2}{-2}=\dfrac{-\left(\sqrt{2}+2\right)^2}{2}=\dfrac{-\left(2+4\sqrt{2}+4\right)}{2}=\dfrac{-6-4\sqrt{2}}{2}=-3-2\sqrt{2}\\ ----------\\ 5,\dfrac{3\sqrt{3}-2\sqrt{3}}{\sqrt{3}-\sqrt{2}}=\dfrac{\sqrt{3}\left(3-2\right)}{\sqrt{3}-\sqrt{2}}=\sqrt{3}\left(\sqrt{3}+\sqrt{2}\right)=\sqrt{9}+\sqrt{6}=3+\sqrt{6}\\ -----------\\ 6,\dfrac{15}{\sqrt{2}-\sqrt{7}}=\dfrac{15\left(\sqrt{2}+\sqrt{7}\right)}{2-7}=-3\left(\sqrt{2}+\sqrt{7}\right)=-3\sqrt{2}-3\sqrt{7}\)