Đáp án:
$\begin{array}{l}
Dkxd:{x^2} - 3x - 10 > 0\\
\Rightarrow \left( {x - 5} \right)\left( {x + 2} \right) > 0\\
\Rightarrow \left[ \begin{array}{l}
x > 5\\
x < - 2
\end{array} \right.\left( 1 \right)\\
\dfrac{{2x - 4}}{{\sqrt {{x^2} - 3x - 10} }} > 1\\
\Rightarrow \dfrac{{2x - 4 - \sqrt {{x^2} - 3x - 10} }}{{\sqrt {{x^2} - 3x - 10} }} > 0\\
\Rightarrow 2x - 3 - \sqrt {{x^2} - 3x - 10} > 0\\
\Rightarrow 2x - 3 > \sqrt {{x^2} - 3x - 10} > 0\\
\Rightarrow \left\{ \begin{array}{l}
2x - 3 > 0\\
{\left( {2x - 3} \right)^2} > {x^2} - 3x - 10
\end{array} \right.\\
\Rightarrow \left\{ \begin{array}{l}
x > \dfrac{3}{2}\\
4{x^2} - 12x + 9 > {x^2} - 3x - 10
\end{array} \right.\\
\Rightarrow \left\{ \begin{array}{l}
x > \dfrac{3}{2}\\
3{x^2} - 9x + 19 > 0\left( {tm} \right)
\end{array} \right.\left( 2 \right)\\
\left( 1 \right);\left( 2 \right) \Leftrightarrow x > 5\\
Vậy\,x > 5
\end{array}$
