Đáp án:

\(\begin{array}{l}
a,\\
x = \dfrac{{144}}{{25}}\\
b,\\
x = 11
\end{array}\)

Giải thích các bước giải:

 Ta có:

\(\begin{array}{l}
a,\\
DKXD:\,\,x \ge 0\\
 - 5x + 7\sqrt x  + 12 = 0\\
 \Leftrightarrow 5x - 7\sqrt x  - 12 = 0\\
 \Leftrightarrow 5{\sqrt x ^2} - 7\sqrt x  - 12 = 0\\
 \Leftrightarrow \left( {5{{\sqrt x }^2} - 12\sqrt x } \right) + \left( {5\sqrt x  - 12} \right) = 0\\
 \Leftrightarrow \sqrt x .\left( {5\sqrt x  - 12} \right) + \left( {5\sqrt x  - 12} \right) = 0\\
 \Leftrightarrow \left( {5\sqrt x  - 12} \right)\left( {\sqrt x  + 1} \right) = 0\\
 \Leftrightarrow \left[ \begin{array}{l}
5\sqrt x  - 12 = 0\\
\sqrt x  + 1 = 0
\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}
\sqrt x  = \dfrac{{12}}{5}\\
\sqrt x  =  - 1
\end{array} \right.\\
\sqrt x  \ge 0 \Rightarrow \sqrt x  = \dfrac{{12}}{5} \Leftrightarrow x = {\left( {\dfrac{{12}}{5}} \right)^2} \Leftrightarrow x = \dfrac{{144}}{{25}}\\
b,\\
DKXD:\,\,\,x \ge  - 5\\
\sqrt {4x + 20}  - 2\sqrt {x + 5}  + \sqrt {9x + 45}  = 12\\
 \Leftrightarrow \sqrt {4.\left( {x + 5} \right)}  - 2\sqrt {x + 5}  + \sqrt {9\left( {x + 5} \right)}  = 12\\
 \Leftrightarrow \sqrt {{2^2}.\left( {x + 5} \right)}  - 2\sqrt {x + 5}  + \sqrt {{3^2}.\left( {x + 5} \right)}  = 12\\
 \Leftrightarrow 2\sqrt {x + 5}  - 2\sqrt {x + 5}  + 3\sqrt {x + 5}  = 12\\
 \Leftrightarrow 3\sqrt {x + 5}  = 12\\
 \Leftrightarrow \sqrt {x + 5}  = 4\\
 \Leftrightarrow x + 5 = {4^2}\\
 \Leftrightarrow x + 5 = 16\\
 \Leftrightarrow x = 11
\end{array}\)