Đáp án:
$\begin{array}{l}
{x^2} - 2\sqrt {{x^2} + 1} - m + 1 > 0\\
\Rightarrow {x^2} + 1 - 2\sqrt {{x^2} + 1} - m > 0\\
Đặt\,\sqrt {{x^2} + 1} = a \ge 1\forall x\\
a){a^2} - 2a - m > 0\forall a \ge 1\\
\Rightarrow {a^2} - 2a > m\forall a \ge 1\\
\Rightarrow GTNN:f\left( a \right) = {a^2} - 2a > m\forall a \ge 1\\
Có: - \frac{b}{{2a}} = 1\\
\Rightarrow GTNN:f\left( a \right) = f\left( 1 \right) = - 1 > m\\
Vậy\,m < - 1\\
b)x \in \left[ {0;\sqrt 3 } \right]\\
\Rightarrow a \in \left[ {1;2} \right]\\
\Rightarrow m < f\left( 2 \right)\\
\Rightarrow m < 0
\end{array}$