Question

(\sqrt(2)-1)/(\sqrt(2)+2)-(2)/(\sqrt(2)+2)-(\sqrt(2)+1)/(\sqrt(2))

Answer 1

đề kiểu gì vậy

(\sqrt(2)-1)/(\sqrt(2)+2)-(2)/(\sqrt(2)+2)-(\sqrt(2)+1)/(\sqrt(2))

ai hiểu đc

Answer 2

\((\sqrt(2)-1)/(\sqrt(2)+2)-(2)/(\sqrt(2)+2)-(\sqrt(2)+1)/(\sqrt(2))\)

đúng không ??

Answer 3

Đề như thế này à

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

Answer 4

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