[tex]a= \sqrt{8}=2 \sqrt{2} \\ b= \frac{ \sqrt{2}-1 }{ \sqrt{2}+1} = \frac{ (\sqrt{2}-1)^{2} }{2-1}=2-2 \sqrt{2}+1=3-2 \sqrt{2} \\ \\ \frac{a+b}{a-b}= \frac{2 \sqrt{2}+3-2 \sqrt{2}}{2 \sqrt{2}-(3-2 \sqrt{2})}= \frac{3}{2 \sqrt{2}-3+2 \sqrt{2}}= \frac{3}{4 \sqrt{2}-3} = \,\, rationalizam\,\,numitorul \\ \\ = \frac{3(4 \sqrt{2}+3)}{16*2-9}= \frac{12\sqrt{2}+9}{32-9}= \frac{12\sqrt{2}+9}{23}[/tex]
[tex]=> \,\,\, \frac{a+b}{a-b} \neq b [/tex]
