Đáp án:
\(\dfrac{{2{x^3} + 54}}{{ - 3{x^3} + 6{x^2} - 18x}}\)
Giải thích các bước giải:
\(\begin{array}{l}
DK:x \ne \left\{ { - 3;0;3} \right\}\\
A = \left( {\dfrac{1}{3} + \dfrac{3}{{{x^2} - 3x}}} \right):\left( {\dfrac{{{x^2}}}{{27 - 3{x^2}}} + \dfrac{1}{{x + 3}}} \right)\\
= \left[ {\dfrac{{{x^2} - 3x + 3.3}}{{3x\left( {x - 3} \right)}}} \right]:\left[ {\dfrac{{{x^2}}}{{2\left( {9 - {x^2}} \right)}} + \dfrac{1}{{x + 3}}} \right]\\
= \left[ {\dfrac{{{x^2} - 3x + 9}}{{3x\left( {x - 3} \right)}}} \right]:\left[ {\dfrac{{{x^2} + 2\left( {3 - x} \right)}}{{2\left( {3 - x} \right)\left( {3 + x} \right)}}} \right]\\
= \left[ {\dfrac{{{x^2} - 3x + 9}}{{3x\left( {x - 3} \right)}}} \right].\dfrac{{2\left( {3 - x} \right)\left( {3 + x} \right)}}{{{x^2} - 2x + 6}}\\
= \dfrac{{{x^2} - 3x + 9}}{{ - 3x}}.\dfrac{{2\left( {x + 3} \right)}}{{{x^2} - 2x + 6}}\\
= \dfrac{{{x^2} - 3x + 9}}{{ - 3x}}.\dfrac{{2x + 6}}{{{x^2} - 2x + 6}}\\
= \dfrac{{2{x^3} + 6{x^2} - 6{x^2} - 18x + 18x + 54}}{{ - 3{x^3} + 6{x^2} - 18x}}\\
= \dfrac{{2{x^3} + 54}}{{ - 3{x^3} + 6{x^2} - 18x}}
\end{array}\)
