Ta có :\(\dfrac{2\left(\sqrt{x}-1\right)}{x\sqrt{x}-1}=\dfrac{2\left(\sqrt{x}-1\right)}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}=\dfrac{2}{x+\sqrt{x}-1}\)
và x = \(3-2\sqrt{2}=\left(\sqrt{2}-1\right)^2\)
Thay x = \(\left(\sqrt{2}-1\right)^2\) vào \(\dfrac{2}{x+\sqrt{x}-1}\) ta có :
\(\dfrac{2}{3-2\sqrt{2}+\sqrt{2}-1-1}=\dfrac{2}{1-\sqrt{2}}\)
Ta có:
\(x=3-2\sqrt{2}=2-2\sqrt{2}+1=\left(\sqrt{2}-1\right)^2\)
\(\dfrac{2\left(\sqrt{x}-1\right)}{x\sqrt{x}-1}=\dfrac{2\left(\sqrt{\left(\sqrt{2}-1\right)^2}-1\right)}{\left(\sqrt{2}-1\right)^2\sqrt{\left(\sqrt{2}-1\right)^2}-1}\)
\(=\dfrac{2\left(\sqrt{2}-1-1\right)}{\left(\sqrt{2}-1\right)^3-1}\)
\(=\dfrac{2\left(\sqrt{2}-2\right)}{\left(\sqrt{2}-2\right)\left(2-2\sqrt{2}+1+\sqrt{2}-1+1\right)}\)
\(=\dfrac{2}{3-\sqrt{2}}=\dfrac{6+2\sqrt{2}}{7}\)
