Đáp án:
$\begin{array}{l}
1)\sqrt {45} + \sqrt {20} - \sqrt 5 \\
= 3\sqrt 5 + 2\sqrt 5 - \sqrt 5 \\
= 4\sqrt 5 \\
2)\dfrac{{x + \sqrt x }}{{\sqrt x }} + \dfrac{{x - 4}}{{\sqrt x + 2}}\\
= \dfrac{{\sqrt x \left( {\sqrt x + 1} \right)}}{{\sqrt x }} + \dfrac{{\left( {\sqrt x + 2} \right)\left( {\sqrt x - 2} \right)}}{{\sqrt x + 2}}\\
= \sqrt x + 1 + \sqrt x - 2\\
= 2\sqrt x - 1\\
2) - {x^2} = kx + 1\\
\Rightarrow {x^2} + kx + 1 = 0\\
\Rightarrow \Delta > 0\\
\Rightarrow {k^2} - 4 > 0\\
\Rightarrow {k^2} > 4\\
\Rightarrow \left[ \begin{array}{l}
k > 2\\
k < - 2
\end{array} \right.\\
3)\\
{x^2} = 4x - {m^2} - 1\\
\Rightarrow {x^2} - 4x + {m^2} + 1 = 0\\
\Rightarrow \Delta ' = 0\\
\Rightarrow 4 - {m^2} - 1 = 0\\
\Rightarrow {m^2} = 3\\
\Rightarrow m = \pm \sqrt 3 \\
Khi:{m^2} = 3\\
\Rightarrow {x^2} - 4x - 4 = 0\\
\Rightarrow {\left( {x - 2} \right)^2} = 0\\
\Rightarrow x = 2\\
\Rightarrow y = {x^2} = 4\\
\Rightarrow Tiep\,diem:\left( {2;4} \right)\\
4){x^2} - 4x + m + 1 = 0\\
\Rightarrow \Delta ' \ge 0\\
\Rightarrow 4 - m - 1 \ge 0\\
\Rightarrow m \le 3\\
Theo\,Viet:\left\{ \begin{array}{l}
{x_1} + {x_2} = 4\\
{x_1}{x_2} = m + 1
\end{array} \right.\\
x_1^2 + x_2^2 = 5\left( {{x_1} + {x_2}} \right)\\
\Rightarrow {\left( {{x_1} + {x_2}} \right)^2} - 2{x_1}{x_2} = 5.4\\
\Rightarrow {4^2} - 2.\left( {m + 1} \right) = 20\\
\Rightarrow m + 1 = - 2\\
\Rightarrow m = - 3\left( {tmdk} \right)
\end{array}$
