Đáp án:
$\begin{array}{l}
a)Dkxd:x \ge 1\\
\sqrt {25x - 25} - \dfrac{{15}}{2}\sqrt {\dfrac{{x - 1}}{9}} = 6 + \sqrt {x - 1} \\
\Leftrightarrow 5.\sqrt {x - 1} - \dfrac{{15}}{2}.\dfrac{{\sqrt {x - 1} }}{3} - \sqrt {x - 1} = 6\\
\Leftrightarrow 5\sqrt {x - 1} - \dfrac{5}{2}\sqrt {x - 1} - \sqrt {x - 1} = 6\\
\Leftrightarrow \dfrac{3}{2}\sqrt {x - 1} = 6\\
\Leftrightarrow \sqrt {x - 1} = 4\\
\Leftrightarrow x - 1 = 16\\
\Leftrightarrow x = 17\left( {tmdk} \right)\\
Vậy\,x = 17\\
b)Dkxd:x \ge 2\\
\sqrt {{x^2} - 4x - 4} = x - 2\\
\Leftrightarrow {x^2} - 4x - 4 = {x^2} - 4x + 4\\
\Leftrightarrow - 4 = 4\left( {ktm} \right)
\end{array}$
Vậy pt vô nghiệm
