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