Question

\(\dfrac{2+\sqrt{2}}{\sqrt{2}+1}+\dfrac{1}{\sqrt{3}+\sqrt{2}}\)

giúp tui vs pls tui đag cần gấp lắm

Answer 1

\(\dfrac{2+\sqrt{2}}{\sqrt{2}+1}+\dfrac{1}{\sqrt{3}+\sqrt{2}}\)

\(=\dfrac{\sqrt{2}\left(\sqrt{2}+1\right)}{\sqrt{2}+1}+\dfrac{\sqrt{3}-\sqrt{2}}{3-2}\)

\(=\sqrt{2}+\sqrt{3}-\sqrt{2}=\sqrt{3}\)