Trebuie sa gasim un binom in acel radical:
[tex]\sqrt{3-2\sqrt{2}}=\sqrt{2-2\sqrt{2}+1}=\sqrt{(\sqrt{2})^2-2\cdot1\cdot\sqrt{2}+1^2}=\\\\
=\sqrt{(\sqrt{2}-1)^2}=|\sqrt{2}-1|=\sqrt{2}-1[/tex]
[tex]a=\sqrt{2}-1-\sqrt{2}=\boxed{-1\ \in Z}[/tex]
Am mai gasit o varianta interesanta cu aceasta formula:
[tex]\sqrt{x\pm\sqrt{y}}=\sqrt{\frac{x+\sqrt{x^2-y}}{2}}\pm\sqrt{\frac{x-\sqrt{x^2-y}}{2}}\\\\
x^2\geq y[/tex]
[tex]\sqrt{3-2\sqrt{2}}=\sqrt{3-\sqrt{8}}=\sqrt{\frac{3+\sqrt{3^2-8}}{2}}-\sqrt{\frac{3-\sqrt{3^2-8}}{2}}=\sqrt{\frac{4}{2}}-\sqrt{\frac{2}{2}}=\\\\
=\sqrt{2}-1[/tex]
Si s-a ajuns la acelasi lucru.
