Đáp án:
$\begin{array}{l}
a)Dkxd:x \ge 1\\
\sqrt {16x - 16} + \sqrt {36x - 36} - 2\sqrt {x - 1} = 16\\
\Leftrightarrow 4\sqrt {x - 1} + 6\sqrt {x - 1} - 2\sqrt {x - 1} = 16\\
\Leftrightarrow 8\sqrt {x - 1} = 16\\
\Leftrightarrow \sqrt {x - 1} = 2\\
\Leftrightarrow \sqrt {x - 1} = 4\\
\Leftrightarrow x - 1 = 16\\
\Leftrightarrow x = 17\left( {tmdk} \right)\\
\text{Vậy}\,x = 17\\
b)M = \sqrt {25{x^2} - 10x + 1} - 3x - 1\\
= \sqrt {{{\left( {5x - 1} \right)}^2}} - 3x - 1\\
= \left| {5x - 1} \right| - 3x - 1\\
Thay\,x = - 2\sqrt 5 \\
\Rightarrow M = \left| {5.\left( { - 2\sqrt 5 } \right) - 1} \right| - 3.\left( { - 2\sqrt 5 } \right) - 1\\
= 10\sqrt 5 + 1 + 6\sqrt 5 - 1\\
= 16\sqrt 5
\end{array}$
