Ta có: (\(3+2\sqrt{2}\)).(\(\sqrt{2}-1\))
= (\(2+2\sqrt{2}+1\)).(\(\sqrt{2}-1\))
= \(\left(\sqrt{2}+1\right)^2\)\(.\left(\sqrt{2}-1\right)\)
= \(\left(\sqrt{2}+1\right).\left(2-1\right)\)
=\(\sqrt{2}+1\)
Suy ra: \(\dfrac{\sqrt{2}+1}{\sqrt{2}-1}=3+2\sqrt{2}\)
ta có
\(\dfrac{\sqrt{2}+1}{\sqrt{2}-1}=3+2\sqrt{2}\)
\(\Leftrightarrow\sqrt{2}+1=\left(3+2\sqrt{2}\right)\cdot\left(\sqrt{2}-1\right)\)
\(\Leftrightarrow\sqrt{2}+1=3\sqrt{2}-3+2\sqrt{2}\cdot\sqrt{2}-2\sqrt{2}\)
\(\Leftrightarrow\sqrt{2}+1=\sqrt{2}+1\)(luôn đúng)
vậy \(\dfrac{\sqrt{2}+1}{\sqrt{2}-1}=3+2\sqrt{2}\)
