Đáp án:
c) \(\begin{array}{l}
f\left( x \right) > 0 \Leftrightarrow x \in \left( { - \dfrac{1}{5};\dfrac{2}{5}} \right)\\
f\left( x \right) < 0 \Leftrightarrow x \in \left( { - \infty ; - \dfrac{1}{5}} \right) \cup \left( {\dfrac{2}{5}; + \infty } \right)
\end{array}\)
Giải thích các bước giải:
\(a)DK:x \ne - \dfrac{1}{2}\)
BXD:
x -∞ -1/2 4/3 +∞
f(x) - // + 0 -
\(\begin{array}{l}
KL:f\left( x \right) > 0 \Leftrightarrow x \in \left( { - \dfrac{1}{2};\dfrac{4}{3}} \right)\\
f\left( x \right) < 0 \Leftrightarrow x \in \left( { - \infty ; - \dfrac{1}{2}} \right) \cup \left( {\dfrac{4}{3}; + \infty } \right)
\end{array}\)
\(b)DK:x \ne \dfrac{1}{5}\)
BXD:
x -∞ 1/5 2/3 +∞
f(x) - // + 0 -
\(\begin{array}{l}
KL:f\left( x \right) > 0 \Leftrightarrow x \in \left( {\dfrac{1}{5};\dfrac{2}{3}} \right)\\
f\left( x \right) < 0 \Leftrightarrow x \in \left( { - \infty ;\dfrac{1}{5}} \right) \cup \left( {\dfrac{2}{3}; + \infty } \right)
\end{array}\)
\(c)DK:x \ne - \dfrac{1}{5}\)
BXD:
x -∞ -1/5 2/5 +∞
f(x) - // + 0 -
\(\begin{array}{l}
KL:f\left( x \right) > 0 \Leftrightarrow x \in \left( { - \dfrac{1}{5};\dfrac{2}{5}} \right)\\
f\left( x \right) < 0 \Leftrightarrow x \in \left( { - \infty ; - \dfrac{1}{5}} \right) \cup \left( {\dfrac{2}{5}; + \infty } \right)
\end{array}\)