Đáp án:

$\begin{array}{l}
a)\sqrt {{{\left( {2x - 1} \right)}^2}}  = 5\\
 \Leftrightarrow \left| {2x - 1} \right| = 5\\
 \Leftrightarrow \left[ \begin{array}{l}
2x - 1 = 5\\
2x - 1 =  - 5
\end{array} \right.\\
 \Leftrightarrow \left[ \begin{array}{l}
2x = 6 \Leftrightarrow x = 3\\
2x =  - 4 \Leftrightarrow x =  - 2
\end{array} \right.\\
Vậy\,x = 3;x =  - 2\\
c)Dkxd:\left\{ \begin{array}{l}
x \ge 3\\
x \le 4
\end{array} \right. \Leftrightarrow 3 \le x \le 4\\
\sqrt {x - 3}  = \sqrt {4 - x} \\
 \Leftrightarrow x - 3 = 4 - x\\
 \Leftrightarrow 2x = 7\\
 \Leftrightarrow x = \dfrac{7}{2}\left( {tmdk} \right)\\
Vậy\,x = \dfrac{7}{2}\\
c)Dkxd:x \ge 0\\
x - 14\sqrt x  + 49 = 0\\
 \Leftrightarrow {\left( {\sqrt x  - 7} \right)^2} = 0\\
 \Leftrightarrow \sqrt x  = 7\\
 \Leftrightarrow x = 49\left( {tmdk} \right)\\
Vậy\,x = 49\\
d)Dkxd:x \ge 1\\
\sqrt {x - 1}  + \sqrt {x + 2}  = 3\\
 \Leftrightarrow x - 1 + 2\sqrt {x - 1} .\sqrt {x + 2}  + x + 2 = 9\\
 \Leftrightarrow 2x + 1 + 2\sqrt {\left( {{x^2} + x - 2} \right)}  = 3\\
 \Leftrightarrow 2\sqrt {{x^2} + x - 2}  = 2 - 2x\left( {dk:x \le 1} \right)\\
Do:\left\{ \begin{array}{l}
x \ge 1\\
x \le 1
\end{array} \right.\\
 + Khi:x = 1\\
 \Leftrightarrow \sqrt {1 - 1}  + \sqrt {1 + 2}  = 3\\
 \Leftrightarrow \sqrt 3  = 3\left( {ktm} \right)\\
Vậy\,x \in \emptyset \\
e)Dkxd:x \ge  - \dfrac{1}{2}\\
\sqrt {{x^2} - x + 9}  = 2x + 1\\
 \Leftrightarrow {x^2} - x + 9 = 4{x^2} + 4x + 1\\
 \Leftrightarrow 3{x^2} + 5x - 8 = 0\\
 \Leftrightarrow 3{x^2} + 8x - 3x - 8 = 0\\
 \Leftrightarrow \left( {3x + 8} \right)\left( {x - 1} \right) = 0\\
 \Leftrightarrow x = 1\left( {do:x \ge  - \dfrac{1}{2}} \right)\\
Vậy\,x = 1\\
f)Dkxd:x \ge  - \dfrac{1}{3}\\
\sqrt {5x + 7}  - \sqrt {x + 3}  = \sqrt {3x + 1} \\
 \Leftrightarrow \sqrt {5x + 7}  = \sqrt {x + 3}  + \sqrt {3x + 1} \\
 \Leftrightarrow 5x + 7 = x + 3 + 2\sqrt {x + 3} .\sqrt {3x + 1}  + 3x + 1\\
 \Leftrightarrow x + 3 = 2\sqrt {\left( {3{x^2} + 10x + 3} \right)} \\
 \Leftrightarrow {x^2} + 6x + 9 = 4\left( {3{x^2} + 10x + 3} \right)\\
 \Leftrightarrow 11{x^2} + 34x + 3 = 0\\
 \Leftrightarrow \left( {x + 3} \right)\left( {11x + 1} \right) = 0\\
 \Leftrightarrow x =  - \dfrac{1}{{11}}\left( {do:x \ge  - \dfrac{1}{3}} \right)\\
Vậy\,x =  - \dfrac{1}{{11}}\\
g){x^2} + 4x + 7 = 4 + x\sqrt {{x^2} + 7} \\
 \Leftrightarrow {x^2} + 7 - x.\sqrt {{x^2} + 7}  + 4x - 4 = 0\\
 \Leftrightarrow {x^2} + 4x + 3 = x\sqrt {{x^2} + 7} \\
 \Leftrightarrow {x^4} + 16{x^2} + 9 + 8{x^3} + 6{x^2} + 24x\\
 = {x^2}\left( {{x^2} + 7} \right)\\
 \Leftrightarrow {x^4} + 8{x^3} + 22{x^2} + 24x + 9 = {x^4} + 7{x^2}\\
 \Leftrightarrow 8{x^3} + 15{x^2} + 24x + 9 = 0\\
 \Leftrightarrow x =  - 0,48\\
Vậy\,x =  - 0,48\\
h)Dkxd:x \ge 0\\
\dfrac{{\sqrt x  - 5}}{{\sqrt x  + 5}} < \dfrac{1}{4}\\
 \Leftrightarrow \dfrac{{4\sqrt x  - 20 - \sqrt x  - 5}}{{4\left( {\sqrt x  + 5} \right)}} < 0\\
 \Leftrightarrow \dfrac{{3\sqrt x  - 25}}{{4\left( {\sqrt x  + 5} \right)}} < 0\\
 \Leftrightarrow 3\sqrt x  - 25 < 0\\
 \Leftrightarrow \sqrt x  < \dfrac{{25}}{3}\\
 \Leftrightarrow x < \dfrac{{625}}{9}\\
Vậy\,0 \le x < \dfrac{{625}}{9}\\
i)Dkxd:x \ge 0;x \ne 1\\
\dfrac{{\sqrt x  + 2}}{{\sqrt x  - 1}} = \dfrac{3}{5}\\
 \Leftrightarrow 5\sqrt x  + 10 = 3\sqrt x  - 3\\
 \Leftrightarrow 2\sqrt x  + 13 = 0\left( {ktm} \right)\\
Vậy\,x \in \emptyset \\
k)Dkxd:x \ge 0\\
\dfrac{{3\sqrt x  - 1}}{{\sqrt x  + 2}} > \dfrac{1}{2}\\
 \Leftrightarrow \dfrac{{6\sqrt x  - 2 - \sqrt x  - 2}}{{2\left( {\sqrt x  + 2} \right)}} > 0\\
 \Leftrightarrow 5\sqrt x  - 4 > 0\\
 \Leftrightarrow \sqrt x  > \dfrac{4}{5}\\
 \Leftrightarrow x > \dfrac{{16}}{{25}}\\
Vậy\,x > \dfrac{{16}}{{25}}
\end{array}$