Đáp án:

\[\begin{array}{l}
2)\\
b)DKxd:3{x^2} + 1 \ge 0\\
 \Leftrightarrow 3{x^2} \ge  - 1\left( {tm} \right)\\
\text{Vậy biểu thức xác định với mọi x}\\
d)Dkxd:{x^2} + 4x + 4 \ge 0\\
 \Leftrightarrow {\left( {x + 2} \right)^2} \ge 0\left( {tm} \right)\\
\text{Vậy biểu thức xác định với mọi x}\\
f)Dkxd: - 4{x^2} + 4x - 1 \ge 0\\
 \Leftrightarrow 4{x^2} - 4x + 1 \le 0\\
 \Leftrightarrow {\left( {2x - 1} \right)^2} \le 0\\
 \Leftrightarrow 2x - 1 = 0\\
 \Leftrightarrow x = \dfrac{1}{2}\\
\text{Vậy}\,x = \dfrac{1}{2}\\
g)Dkxd:x\left( {x + 2} \right) \ge 0\\
 \Leftrightarrow \left[ \begin{array}{l}
\left\{ \begin{array}{l}
x \ge 0\\
x + 2 \ge 0
\end{array} \right.\\
\left\{ \begin{array}{l}
x \le 0\\
x + 2 \le 0
\end{array} \right.
\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}
\left\{ \begin{array}{l}
x \ge 0\\
x \ge  - 2
\end{array} \right.\\
\left\{ \begin{array}{l}
x \le 0\\
x \le  - 2
\end{array} \right.
\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}
x \ge 0\\
x \le  - 2
\end{array} \right.\\
\text{Vậy}\,x \ge 0\,hoac\,x \le  - 2\\
h)4 - {x^2} \ge 0\\
 \Leftrightarrow {x^2} \le 4\\
 \Leftrightarrow  - 2 \le x \le 2\\
\text{Vậy}\, - 2 \le x \le 2\\
i)Dkxd:\left\{ \begin{array}{l}
2 - x \ge 0\\
x + 3 < 0
\end{array} \right.\\
 \Leftrightarrow \left\{ \begin{array}{l}
x \le 2\\
x <  - 3
\end{array} \right.\\
 \Leftrightarrow x \le 2\\
\text{Vậy}\,x \le 2\\
B3)\\
4)x + 5\sqrt x  + 4\\
 = {\left( {\sqrt x } \right)^2} + 5\sqrt x  + 4\\
 = \left( {\sqrt x  + 1} \right)\left( {\sqrt x  + 4} \right)\\
7)x - 5\sqrt x  + 6\\
 = \left( {\sqrt x  - 2} \right)\left( {\sqrt x  - 3} \right)\\
8)\sqrt x  - x\\
 = \sqrt x \left( {1 - \sqrt x } \right)\\
10)x + 2\sqrt {x - 1} \\
 = x - 1 + 2\sqrt {x - 1}  + 1\\
 = {\left( {\sqrt {x - 1}  + 1} \right)^2}\\
11)x + 1 + 4\sqrt {x - 3} \\
 = x - 3 + 4\sqrt {x - 3}  + 4\\
 = {\left( {\sqrt {x - 3}  + 2} \right)^2}\\
B4)\\
b)\dfrac{{3x - 6\sqrt x  - 9}}{{x - \sqrt x }}\\
 = \dfrac{{3\left( {x - 2\sqrt x  - 3} \right)}}{{\sqrt x \left( {\sqrt x  - 1} \right)}}\\
 = \dfrac{{3\left( {\sqrt x  + 1} \right)\left( {\sqrt x  - 3} \right)}}{{\sqrt x \left( {\sqrt x  - 1} \right)}}\\
c)\dfrac{{x\sqrt x  - 8 + 2x - 4\sqrt x }}{{x - 4}}\\
 = \dfrac{{\left( {\sqrt x  - 2} \right)\left( {x + 2\sqrt x  + 4} \right) + 2\sqrt x \left( {\sqrt x  - 2} \right)}}{{\left( {\sqrt x  - 2} \right)\left( {\sqrt x  + 2} \right)}}\\
 = \dfrac{{\left( {\sqrt x  - 2} \right)\left( {x + 2\sqrt x  + 4 + 2\sqrt x } \right)}}{{\left( {\sqrt x  - 2} \right)\left( {\sqrt x  + 2} \right)}}\\
 = \dfrac{{x + 4\sqrt x  + 4}}{{\sqrt x  + 2}}\\
 = \dfrac{{{{\left( {\sqrt x  + 2} \right)}^2}}}{{\sqrt x  + 2}}\\
 = \sqrt x  + 2\\
B5)\\
4)2 = 1 + 1 < 1 + \sqrt 2 \\
7)5\sqrt 6  = \sqrt {25.6}  = \sqrt {150} \\
6\sqrt 5  = \sqrt {36.5}  = \sqrt {180} \\
 \Leftrightarrow 5\sqrt 6  < 6\sqrt 5 \\
8) - 5\sqrt 7  =  - \sqrt {{5^2}.7}  =  - \sqrt {175} \\
 - 7\sqrt 5  =  - \sqrt {59.5}  =  - \sqrt {295} \\
 \Leftrightarrow  - 5\sqrt 7  >  - 7\sqrt 5 \\
B6)\\
a)\sqrt {2x - 1}  = \sqrt 5 \\
 \Leftrightarrow 2x - 1 = 5\\
 \Leftrightarrow 2x = 6\\
 \Leftrightarrow x = 3\\
\text{Vậy}\,x = 3\\
b)\sqrt {x - 5}  = 3\\
 \Leftrightarrow x - 5 = 9\\
 \Leftrightarrow x = 14\\
\text{Vậy}\,x = 14\\
c)\sqrt {{{\left( {x - 3} \right)}^2}}  = 9\\
 \Leftrightarrow {\left( {x - 3} \right)^2} = 81\\
 \Leftrightarrow \left[ \begin{array}{l}
x - 3 = 9\\
x - 3 =  - 9
\end{array} \right.\\
 \Leftrightarrow \left[ \begin{array}{l}
x = 12\\
x =  - 6
\end{array} \right.\\
\text{Vậy}\,x = 12;x =  - 6\\
d)\sqrt {4{x^2} - 4x + 1}  = 3\\
 \Leftrightarrow 4{x^2} - 4x + 1 = 9\\
 \Leftrightarrow 4{x^2} - 4x - 8 = 0\\
 \Leftrightarrow {x^2} - x - 2 = 0\\
 \Leftrightarrow \left( {x - 2} \right)\left( {x + 1} \right) = 0\\
 \Leftrightarrow x = 2;x =  - 1\\
\text{Vậy}\,x = 2;x =  - 1\\
e)\sqrt {{{\left( {x + 3} \right)}^2}}  = x + 3\\
 \Leftrightarrow \left| {x + 3} \right| = x + 3\\
 \Leftrightarrow x + 3 \ge 0\\
 \Leftrightarrow x \ge  - 3\\
\text{Vậy}\,x \ge  - 3\\
f)\left( {x + 2} \right)\sqrt {x - 3}  = 0\\
Dkxd:x \ge 3\\
 \Leftrightarrow \left[ \begin{array}{l}
x + 2 = 0\\
\sqrt {x - 3}  = 0
\end{array} \right.\\
 \Leftrightarrow \left[ \begin{array}{l}
x =  - 2\left( {ktm} \right)\\
x = 3\left( {tm} \right)
\end{array} \right.\\
\text{Vậy}\,x = 3
\end{array}\]