Đáp án:
b) \(\begin{array}{l}
f\left( x \right) > 0 \Leftrightarrow x \in \left( {\dfrac{1}{4};\dfrac{3}{2}} \right)\\
f\left( x \right) < 0 \Leftrightarrow x \in \left( { - \infty ;\dfrac{1}{4}} \right) \cup \left( {\dfrac{3}{2}; + \infty } \right)
\end{array}\)
Giải thích các bước giải:
\(\begin{array}{l}
a)DK:x \ne \dfrac{1}{3}\\
Xét:f\left( x \right) = 0\\
\to \dfrac{{\left( {x - 2} \right)\left( {x + 1} \right)}}{{3x - 1}} = 0\\
\to \left[ \begin{array}{l}
x = 2\\
x = - 1
\end{array} \right.
\end{array}\)
BXD:
x -∞ -1 1/3 2 +∞
f(x) - 0 + // - 0 +
\(\begin{array}{l}
KL:f\left( x \right) > 0 \Leftrightarrow x \in \left( { - 1;\dfrac{1}{3}} \right) \cup \left( {2; + \infty } \right)\\
f\left( x \right) < 0 \Leftrightarrow x \in \left( { - \infty ; - 1} \right) \cup \left( {\dfrac{1}{3};2} \right)
\end{array}\)
\(b)DK:x \ne \dfrac{3}{2}\)
BXD:
x -∞ 1/4 3/2 +∞
f(x) - 0 + // -
\(\begin{array}{l}
KL:f\left( x \right) > 0 \Leftrightarrow x \in \left( {\dfrac{1}{4};\dfrac{3}{2}} \right)\\
f\left( x \right) < 0 \Leftrightarrow x \in \left( { - \infty ;\dfrac{1}{4}} \right) \cup \left( {\dfrac{3}{2}; + \infty } \right)
\end{array}\)
\(c)DK:x \ne \dfrac{4}{5}\)
BXD:
x -∞ 0 4/5 3 +∞
f(x) + 0 - // + 0 -
\(\begin{array}{l}
d)DK:x \ne \left\{ { - \dfrac{1}{3};2} \right\}\\
f\left( x \right) = - \dfrac{4}{{3x + 1}} - \dfrac{3}{{2 - x}}\\
= \dfrac{{ - 8 + 4x - 9x - 3}}{{\left( {2 - x} \right)\left( {3x + 1} \right)}}\\
= \dfrac{{ - 11 - 5x}}{{\left( {2 - x} \right)\left( {3x + 1} \right)}}
\end{array}\)
BXD:
x -∞ -11/5 -1/3 2 +∞
f(x) - 0 + // - // +
\(e)DK:x \ne \left\{ { - \dfrac{1}{2}; - \dfrac{1}{3}} \right\}\)
BXD:
x -∞ -1/2 -1/3 5/2 +∞
f(x) + // - // + 0 -
