Ta có \(\dfrac{1}{\sqrt{2}-1}-\dfrac{1}{\sqrt{2}+1}=\dfrac{\sqrt{2}+1}{\left(\sqrt{2}-1\right)\left(\sqrt{2}+1\right)}-\dfrac{\sqrt{2}-1}{\left(\sqrt{2}-1\right)\left(\sqrt{2}+1\right)}=\dfrac{\sqrt{2}+1-\sqrt{2}+1}{\left(\sqrt{2}-1\right)\left(\sqrt{2}+1\right)}=\dfrac{2}{2-1}=2\)
\(\sqrt{\left(1-2\sqrt{2}\right)^2}=\left|1-2\sqrt{2}\right|=2\sqrt{2}-1\)
Ta có \(8< 9\Leftrightarrow\sqrt{8}< \sqrt{9}\Leftrightarrow2\sqrt{2}< 3\Leftrightarrow2\sqrt{2}< 2+1\Leftrightarrow2\sqrt{2}-1< 2\Leftrightarrow\sqrt{\left(1-2\sqrt{2}\right)^2}< \dfrac{1}{\sqrt{2}-1}-\dfrac{1}{\sqrt{2}+1}\)
Vậy \(\dfrac{1}{\sqrt{2}-1}-\dfrac{1}{\sqrt{2}+1}>\sqrt{\left(1-2\sqrt{2}\right)^2}\)
