Đáp án:

$\begin{array}{l}
1)N = 1 + \sqrt {81}  = 1 + 9 = 10\\
H = \sqrt {{{\left( {3 - \sqrt 5 } \right)}^2}}  + \sqrt 5 \\
 = 3 - \sqrt 5  + \sqrt 5  = 3\\
2)Dkxd:x \ge 0;x \ne 1\\
G = \dfrac{{x - \sqrt x }}{{\sqrt x  - 1}} - \dfrac{{x - 1}}{{\sqrt x  + 1}}\\
 = \dfrac{{\sqrt x \left( {\sqrt x  - 1} \right)}}{{\sqrt x  - 1}} - \dfrac{{\left( {\sqrt x  + 1} \right)\left( {\sqrt x  - 1} \right)}}{{\sqrt x  + 1}}\\
 = \sqrt x  - \left( {\sqrt x  - 1} \right)\\
 = 1\\
B2:\\
1)b)d' \bot d\\
 \Rightarrow d':y =  - \dfrac{1}{3}x + b\\
Xét: - {x^2} =  - \dfrac{1}{3}x + b\\
 \Rightarrow {x^2} - \dfrac{1}{3}x + b = 0\\
 \Rightarrow \Delta  = 0\\
 \Rightarrow {\left( {\dfrac{1}{3}} \right)^2} - 4b = 0\\
 \Rightarrow 4b = \dfrac{1}{9}\\
 \Rightarrow b = \dfrac{1}{{36}}\\
 \Rightarrow d':y =  - \dfrac{1}{3}x + \dfrac{1}{{36}}\\
2)\left\{ \begin{array}{l}
3x - y = 5\\
5x + 2y = 23
\end{array} \right.\\
 \Rightarrow \left\{ \begin{array}{l}
6x - 2y = 10\\
5x + 2y = 23
\end{array} \right.\\
 \Rightarrow \left\{ \begin{array}{l}
11x = 33\\
3x - y = 5
\end{array} \right.\\
 \Rightarrow \left\{ \begin{array}{l}
x = 3\\
y = 3x - 5 = 4
\end{array} \right.\\
Vậy\,\left( {x;y} \right) = \left( {3;4} \right)
\end{array}$