Đáp án:
a) \(\begin{array}{l}
f\left( x \right) > 0 \Leftrightarrow x \in \left( { - \infty ; - 3} \right) \cup \left( {0;3} \right)\\
f\left( x \right) < 0 \Leftrightarrow x \in \left( { - 3;0} \right) \cup \left( {3; + \infty } \right)
\end{array}\)
Giải thích các bước giải:
\(\begin{array}{l}
a)DK:x \ne \pm 3\\
\dfrac{1}{{3 - x}} - \dfrac{1}{{3 + x}} = \dfrac{{x + 3 - 3 + x}}{{\left( {3 - x} \right)\left( {x + 3} \right)}}\\
= \dfrac{{2x}}{{\left( {3 - x} \right)\left( {x + 3} \right)}}
\end{array}\)
BXD:
x -∞ -3 0 3 +∞
f(x) + // - 0 + // -
\(\begin{array}{l}
KL:f\left( x \right) > 0 \Leftrightarrow x \in \left( { - \infty ; - 3} \right) \cup \left( {0;3} \right)\\
f\left( x \right) < 0 \Leftrightarrow x \in \left( { - 3;0} \right) \cup \left( {3; + \infty } \right)
\end{array}\)
\(\begin{array}{l}
b)DK:x \ne \pm 2\\
\dfrac{{x - 2 + x + 2}}{{\left( {x - 2} \right)\left( {x + 2} \right)}} = \dfrac{{2x}}{{\left( {x + 2} \right)\left( {x - 2} \right)}}
\end{array}\)
BXD:
x -∞ -2 0 2 +∞
f(x) - // + 0 - // +
\(\begin{array}{l}
c)DK:x \ne \pm 2\\
\dfrac{{x - 2 - x - 2}}{{\left( {x - 2} \right)\left( {x + 2} \right)}} = \dfrac{{ - 4}}{{\left( {x + 2} \right)\left( {x - 2} \right)}}
\end{array}\)
BXD:
x -∞ -2 2 +∞
f(x) - // + // -
