Question

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

Answer 1

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

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

Answer 2

giup mình với mình cần gấp

 

Answer 3

Ta có: \(\dfrac{\sqrt{2}-1}{2+\sqrt{2}}-\dfrac{2}{2+\sqrt{2}}+\dfrac{\sqrt{2}+1}{\sqrt{2}}\)

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

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