\(\begin{align}
  & a=-1\pm \sqrt{2} \\
& {{b}^{2}}=1+\sqrt{2} \\
& a{{b}^{2}}\ne 1,a\ne -1+\sqrt{2} \\
&  \\
& \frac{a{{b}^{2}}+{{b}^{2}}+1}{a}={{b}^{2}}+\frac{{{b}^{2}}+1}{a}=1+\sqrt{2}+\frac{2+\sqrt{2}}{-1-\sqrt{2}}=1 \\
\end{align}\)

題目的確沒說b是實數
不過\({{b}^{2}}=1-\sqrt{2},a=-1+\sqrt{2}\)代入一樣可求出答案