Вопрос в картинках...

$\sqrt{x-6}\cdot \sqrt{x-12}\ \textless \ x-1$

$\left\{\begin{matrix} x-1 &\ \textgreater \ &0 \\ (x-6)(x-12) &\ \textless \ &(x-1)^2 \\ x-6 &\geqslant &0 \\ x-12 &\geqslant &0 \end{matrix}\right.$

$x-1\ \textgreater \ 0\\ x\ \textgreater \ 1\\ x\in(1;+\infty)$

$(x-6)(x-12)\ \textless \ (x-1)^2\\ (x-6)(x-12)-(x-1)^2\ \textless \ 0\\ x^2-18x+72-x^2+2x-1\ \textless \ 0\\ -16x\ \textless \ -71\\\\ x\ \textgreater \ \frac{71}{16}\\\\ x\in(\frac{71}{16};+\infty)$

$x-6\geqslant 0\\ x\geqslant 6\\ x\in[6;+\infty)\\\\\\ x-12\geqslant 0\\ x\geqslant 12\\ x\in(12;+\infty)$

Ответ: $x\in[12;+\infty)$