Đáp án:
$\begin{array}{l}
a)\sqrt {64} - \sqrt x > 2\sqrt 4 + \dfrac{{\sqrt {625} }}{{\sqrt {25} }}\\
\Leftrightarrow 8 - \sqrt x > 2.2 + \sqrt {25} \\
\Leftrightarrow 8 - \sqrt x > 4 + 5\\
\Leftrightarrow \sqrt x < 8 - 9\\
\Leftrightarrow \sqrt x < - 1\left( {ktm} \right)\\
Vậy\,x \in \emptyset \\
b)25 + \sqrt {16} + 4\sqrt x = \sqrt 9 + \sqrt {196} :\sqrt {49} \\
\Leftrightarrow 25 + 4 + 4\sqrt x = 3 + \sqrt 4 \\
\Leftrightarrow 29 + 4\sqrt x = 3 + 2\\
\Leftrightarrow 29 + 4\sqrt x = 5\\
\Leftrightarrow 4\sqrt x = - 24\\
\Leftrightarrow \sqrt x = - 6\left( {ktm} \right)\\
Vậy\,x \in \emptyset
\end{array}$