Đáp án:

$\begin{array}{l}
1)Dkxd:\\
a)x + 2 \ge 0\\
 \Rightarrow x \ge  - 2\\
b)10 - 5x \ge 0\\
 \Rightarrow 5x \le 10\\
 \Rightarrow x \le 2\\
c)x - 2010 > 0\\
 \Rightarrow x > 2010\\
2)P = \sqrt {{x^2} + 6x + 2011} \\
 = \sqrt {{x^2} + 6x + 9 + 2002} \\
 = \sqrt {{{\left( {x + 3} \right)}^2} + 2002}  \ge \sqrt {2002} \\
 \Rightarrow GTNN:P = \sqrt {2002} \,khi:x =  - 3\\
3)\sqrt {2012}  - \sqrt {2011} \\
 = \dfrac{{\left( {\sqrt {2012}  - \sqrt {2011} } \right)\left( {\sqrt {2012}  + \sqrt {2011} } \right)}}{{\sqrt {2012}  + \sqrt {2011} }}\\
 = \dfrac{{2012 - 2011}}{{\sqrt {2012}  + \sqrt {2011} }}\\
 = \dfrac{1}{{\sqrt {2012}  + \sqrt {2011} }}\\
\sqrt {2011}  - \sqrt {2010} \\
 = \dfrac{{\left( {\sqrt {2011}  - \sqrt {2010} } \right)\left( {\sqrt {2011}  + \sqrt {2010} } \right)}}{{\sqrt {2011}  + \sqrt {2010} }}\\
 = \dfrac{{2011 - 2010}}{{\sqrt {2011}  + \sqrt {2010} }}\\
 = \dfrac{1}{{\sqrt {2011}  + \sqrt {2010} }}\\
Do:\sqrt {2012}  + \sqrt {2011}  > \sqrt {2011}  + \sqrt {2010} \\
 \Rightarrow \dfrac{1}{{\sqrt {2012}  + \sqrt {2011} }} < \dfrac{1}{{\sqrt {2011}  + \sqrt {2010} }}\\
 \Rightarrow \sqrt {2012}  - \sqrt {2011}  < \sqrt {2011}  - \sqrt {2010} \\
4)a)\sqrt {4{x^2} - 12x + 9}  = 7\\
 \Rightarrow \sqrt {{{\left( {2x - 3} \right)}^2}}  = 7\\
 \Rightarrow \left| {2x - 3} \right| = 7\\
 \Rightarrow \left[ \begin{array}{l}
2x - 3 = 7\\
2x - 3 =  - 7
\end{array} \right.\\
 \Rightarrow \left[ \begin{array}{l}
x = 5\\
x =  - 2
\end{array} \right.\\
b)Dkxd:x \ge 1\\
5\sqrt {9x - 9}  - \sqrt {4x - 4}  - \sqrt {x - 1}  = 36\\
 \Rightarrow 15\sqrt {x - 1}  - 2\sqrt {x - 1}  - \sqrt {x - 1}  = 36\\
 \Rightarrow 12\sqrt {x - 1}  = 36\\
 \Rightarrow \sqrt {x - 1}  = 3\\
 \Rightarrow x - 1 = 9\\
 \Rightarrow x = 10\left( {tmdk} \right)
\end{array}$