\(\dfrac{{x\,\left| {\,x-3\,} \right| + {x^2}-9}}{{2{x^3}-3{x^2}-9x}}\)
\(2{x^3}-3{x^2}-9x \ne 0\,\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\,\,x\, \in \,\left( {-\infty ;-\dfrac{3}{2}} \right) \cup \left( {-\dfrac{3}{2};0} \right) \cup \left( {0;3} \right) \cup \left( {3;\infty } \right).\)
Если \(x\, \in \,\left( {-\infty ;-\dfrac{3}{2}} \right) \cup \left( {-\dfrac{3}{2};0} \right) \cup \left( {0;3} \right)\), то \(\left| {x-3} \right| = -x + 3\):
\(\dfrac{{x\left| {x-3} \right| + {x^2}-9}}{{2{x^3}-3{x^2}-9x}} = \dfrac{{-{x^2} + 3x + {x^2}-9}}{{x\left( {2{x^2}-3x-9} \right)}} = \dfrac{{3x-9}}{{x\left( {x-3} \right)\left( {2x + 3} \right)}} = \)
\( = \dfrac{{3\left( {x-3} \right)}}{{x\left( {x-3} \right)\left( {2x + 3} \right)}} = \dfrac{3}{{x\left( {2x + 3} \right)}}.\)
Если \(x\, \in \,\left( {3;\infty } \right)\), то \(\left| {x-3} \right| = x-3\):
\(\dfrac{{x\left| {x-3} \right| + {x^2}-9}}{{2{x^3}-3{x^2}-9x}} = \dfrac{{{x^2}-3x + {x^2}-9}}{{x\left( {2{x^2}-3x-9} \right)}} = \dfrac{{2{x^2}-3x-9}}{{x\left( {2{x^2}-3x-9} \right)}} = \dfrac{1}{x}.\)
Ответ: \(\dfrac{3}{{x\left( {2x + 3} \right)}},\) если \(x \in \left( {-\infty ;-\dfrac{3}{2}} \right) \cup \left( {-\dfrac{3}{2};0} \right) \cup \left( {0;3} \right);\) \(\dfrac{1}{x},\) если \(x \in \left( {3; + \infty } \right).\)