Đáp án:
c. x<-3
Giải thích các bước giải:
\(\begin{array}{l}
a.DK:x \ne \pm 3\\
A = \left[ {\dfrac{{2{x^2} - 6x + {x^2} + 3x - 3{x^2} - 3}}{{\left( {x - 3} \right)\left( {x + 3} \right)}}} \right]:\left( {\dfrac{{2x - 2 - x + 3}}{{x - 3}}} \right)\\
= \dfrac{{ - 3x - 3}}{{\left( {x - 3} \right)\left( {x + 3} \right)}}.\dfrac{{x - 3}}{{x + 1}}\\
= \dfrac{{ - 3}}{{x + 3}}\\
b.\left| {x + 1} \right| = 2\\
\to \left[ \begin{array}{l}
x + 1 = 2\\
x + 1 = - 2
\end{array} \right. \to \left[ \begin{array}{l}
x = 1\\
x = - 3\left( l \right)
\end{array} \right.\\
\to A = \dfrac{{ - 3}}{{1 + 3}} = - \dfrac{3}{4}\\
c.A > 0\\
\to \dfrac{{ - 3}}{{x + 3}} > 0\\
\to x + 3 < 0\\
\to x < - 3
\end{array}\)
