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