Đáp án:
$\begin{array}{l}
b)Dkxd:a \ge 0;a \ne 1\\
M = - {a^2}\\
\Rightarrow a - 1 = - {a^2}\\
\Rightarrow {a^2} + a - 1 = 0\\
\Rightarrow {a^2} + 2.a.\dfrac{1}{2} + \dfrac{1}{4} = \dfrac{5}{4}\\
\Rightarrow {\left( {a + \dfrac{1}{2}} \right)^2} = {\left( {\dfrac{{\sqrt 5 }}{2}} \right)^2}\\
\Rightarrow \left[ \begin{array}{l}
a = \dfrac{{ - \sqrt 5 - 1}}{2}\left( {ktm} \right)\\
a = \dfrac{{\sqrt 5 - 1}}{2}\left( {tm} \right)
\end{array} \right.\\
Vậy\,a = \dfrac{{\sqrt 5 - 1}}{2}
\end{array}$
