Đáp án:
$\begin{array}{l}
1){\left( {\sqrt x - 1} \right)^2} = t\left( {t \ge 0} \right)\\
\Rightarrow 9{t^2} + 5t - 4 = 0\\
\Rightarrow 9{t^2} + 9t - 4t - 4 = 0\\
\Rightarrow \left( {t + 1} \right)\left( {9t - 4} \right) = 0\\
\Rightarrow \left[ \begin{array}{l}
t = - 1\left( {ktm} \right)\\
t = \frac{4}{9}\left( {tm} \right)
\end{array} \right.\\
\Rightarrow {\left( {\sqrt x - 1} \right)^2} = \frac{4}{9}\\
\Rightarrow \left[ \begin{array}{l}
\sqrt x - 1 = \frac{2}{3}\\
\sqrt x - 1 = - \frac{2}{3}
\end{array} \right.\\
\Rightarrow \left[ \begin{array}{l}
\sqrt x = \frac{5}{3} \Rightarrow x = \frac{{25}}{9}\\
\sqrt x = \frac{1}{3} \Rightarrow x = \frac{1}{9}
\end{array} \right.\\
2)a)\left( { - 1;0} \right) \in d\\
\Rightarrow 0 = 2.\left( { - 1} \right) - m + 5\\
\Rightarrow m = 3\\
b){x^2} = 2x - m + 5\\
\Rightarrow {x^2} - 2x + m - 5 = 0\\
\Rightarrow \Delta ' > 0\\
\Rightarrow 1 - m + 5 > 0\\
\Rightarrow m < 6\\
Theo\,Viet:\left\{ \begin{array}{l}
{x_1} + {x_2} = 2\\
{x_1}{x_2} = m - 5
\end{array} \right.\\
x_1^2 - 2{x_1} + m - 5 = 0\\
\Rightarrow x_1^2 = 2{x_1} - m + 5\\
Do:x_1^2 + 2{x_2} < 6\\
\Rightarrow 2{x_1} - m + 5 + 2{x_2} < 6\\
\Rightarrow 2\left( {{x_1} + {x_2}} \right) - m + 5 - 6 < 0\\
\Rightarrow 2.2 - m - 1 < 0\\
\Rightarrow m > 3\\
Vậy\,3 < m < 6
\end{array}$
