\[\begin{array}{l}
{\rm{ }}m{x^2} + x + m + 1 = 0\\
Voi:m = 0 = > 0.{x^2} + x + 0 + 1 = 0\\
= > x + 1 = 0\\
= > x = - 1\left( {ktm} \right)\\
Voi:m \ne 0\\
\Delta = {1^2} - 4.m\left( {m + 1} \right) = 1 - 4{m^2} - 4m
\end{array}\]

Để pt có nghiệm kép thì 

\[\begin{array}{l}
\Delta = 0 = > 1 - 4{m^2} - 4m = 0\\
\Leftrightarrow - 4{m^2} - 4m - 1 + 2 = 0\\
\Leftrightarrow - {\left( {2m + 1} \right)^2} + 2 = 0\\
\Leftrightarrow {\left( {2m + 1} \right)^2} = 2\\
\Leftrightarrow \left[ \begin{array}{l}
2m + 1 = \sqrt 2 \\
2m + 1 = - \sqrt 2
\end{array} \right.\\
\Leftrightarrow \left[ \begin{array}{l}
m = \frac{{\sqrt 2 - 1}}{2}\\
m = \frac{{ - \sqrt 2 - 1}}{2}
\end{array} \right.
\end{array}\]

Khi đó nghiệm kép là: 

\[\begin{array}{l}
x = \frac{{ - b}}{{2{\rm{a}}}}\\
+ )m = \frac{{\sqrt 2 - 1}}{2} = > x = \frac{{ - 1}}{{2m}} = \frac{{ - 1}}{{\frac{{\sqrt 2 - 1}}{2}}} = - 2 - 2\sqrt 2 \\
= )m = \frac{{ - \sqrt 2 - 1}}{2} = > x = \frac{{ - 1}}{{2m}} = \frac{{ - 1}}{{\frac{{ - \sqrt 2 - 1}}{2}}} = - 2 + 2\sqrt 2
\end{array}\]