Đáp án:
$\begin{array}{l}
y = {x^2} + \left( {2 - a} \right).x - 1 + a > 0\forall x < 1\\
\Rightarrow {x^2} + 2x - 1 + \left( {1 - x} \right).a > 0\forall x < 1\\
\Rightarrow {x^2} + 2x - 1 > \left( {x - 1} \right).a\forall x < 1\\
\Rightarrow \dfrac{{{x^2} + 2x - 1}}{{x - 1}} < a\left( {do:x < 1} \right)\\
Dat:g\left( x \right) = \dfrac{{{x^2} + 2x - 1}}{{x - 1}}\\
\Rightarrow g\left( x \right) < a\forall x < 1\\
\Rightarrow \mathop {max}\limits_{\left( { - \infty ;1} \right)} g\left( x \right) < a\\
g'\left( x \right) = \dfrac{{\left( {2x + 2} \right).\left( {x - 1} \right) - {x^2} - 2x + 1}}{{{{\left( {x - 1} \right)}^2}}}\\
= \dfrac{{2{x^2} - 2 - {x^2} - 2x + 1}}{{{{\left( {x - 1} \right)}^2}}}\\
= \dfrac{{{x^2} - 2x - 1}}{{{{\left( {x - 1} \right)}^2}}} = 0\\
\Rightarrow {x^2} - 2x - 1 = 0\\
\Rightarrow \left[ \begin{array}{l}
x = 1 + \sqrt 2 \\
x = 1 - \sqrt 2
\end{array} \right.\\
\Rightarrow \mathop {max}\limits_{\left( { - \infty ;1} \right)} g\left( x \right) = g\left( {1 - \sqrt 2 } \right) = 4 - 2\sqrt 2 \\
\Rightarrow a > 4 - 2\sqrt 2
\end{array}$
