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