Đáp án:
b) A=-1
Giải thích các bước giải:
\(\begin{array}{l}
a)DK:x \ne \left\{ { - 6;0;6} \right\}\\
b)A = \left( {\dfrac{x}{{{x^2} - 36}} - \dfrac{{x - 6}}{{{x^2} + 6x}}} \right):\dfrac{{2x - 6}}{{{x^2} + 6x}} + \dfrac{x}{{6 - x}}\\
= \left[ {\dfrac{{{x^2} - {{\left( {x - 6} \right)}^2}}}{{x\left( {x - 6} \right)\left( {x + 6} \right)}}} \right].\dfrac{{x\left( {x + 6} \right)}}{{2x - 6}} + \dfrac{x}{{6 - x}}\\
= \dfrac{{{x^2} - {x^2} + 12x - 36}}{{x\left( {x - 6} \right)\left( {x + 6} \right)}}.\dfrac{{x\left( {x + 6} \right)}}{{2\left( {x - 3} \right)}} + \dfrac{x}{{6 - x}}\\
= \dfrac{{12\left( {x - 3} \right)}}{{x\left( {x - 6} \right)\left( {x + 6} \right)}}.\dfrac{{x\left( {x + 6} \right)}}{{2\left( {x - 3} \right)}} + \dfrac{x}{{6 - x}}\\
= \dfrac{6}{{x - 6}} + \dfrac{x}{{6 - x}} = \dfrac{{6 - x}}{{x - 6}} = - 1
\end{array}\)
