Ta có:
\(4a^2+a\sqrt{2}-\sqrt{2}=0\)
\(\Leftrightarrow2\sqrt{2}a^2+a-1=0\)
\(\Leftrightarrow a+1=2-2\sqrt{2}a^2\) thế vô ta được
\(\frac{a+1}{\sqrt{a^4+a+1}-a^2}=\frac{2-2\sqrt{2}a^2}{\sqrt{a^4+2-2\sqrt{2}a^2}-a^2}\)
\(=\frac{2-2\sqrt{2}a^2}{\sqrt{\left(\sqrt{2}-a^2\right)^2}-a^2}=\frac{\sqrt{2}\left(\sqrt{2}-2a^2\right)}{\sqrt{2}-2a^2}=\sqrt{2}\)