Đáp án:
a) \(\begin{array}{l}
f\left( x \right) > 0 \Leftrightarrow x \in \left( { - \infty ; - 12} \right) \cup \left( { - 4; - 3} \right) \cup \left( {0; + \infty } \right)\\
f\left( x \right) < 0 \Leftrightarrow x \in \left( { - 12; - 4} \right) \cup \left( { - 3;0} \right)
\end{array}\)
Giải thích các bước giải:
\(\begin{array}{l}
a)DK:x \ne \left\{ { - 4; - 3;0} \right\}\\
f\left( x \right) = \dfrac{{\left( {x + 4} \right)\left( {x + 3} \right) + 2x\left( {x + 3} \right) - 3x\left( {x + 4} \right)}}{{x\left( {x + 3} \right)\left( {x + 4} \right)}}\\
= \dfrac{{{x^2} + 7x + 12 + 2{x^2} + 6x - 3{x^2} - 12x}}{{x\left( {x + 3} \right)\left( {x + 4} \right)}}\\
= \dfrac{{x + 12}}{{x\left( {x + 3} \right)\left( {x + 4} \right)}}
\end{array}\)
BXD:
x -∞ -12 -4 -3 0 +∞
f(x) + 0 - // + // - // +
\(\begin{array}{l}
KL:f\left( x \right) > 0 \Leftrightarrow x \in \left( { - \infty ; - 12} \right) \cup \left( { - 4; - 3} \right) \cup \left( {0; + \infty } \right)\\
f\left( x \right) < 0 \Leftrightarrow x \in \left( { - 12; - 4} \right) \cup \left( { - 3;0} \right)
\end{array}\)
\(\begin{array}{l}
b)DK:x \ne \pm 1\\
g\left( x \right) = \dfrac{{{x^2} - 1 - {x^2} + 3x - 1}}{{\left( {x - 1} \right)\left( {x + 1} \right)}}\\
= \dfrac{{3x - 2}}{{\left( {x - 1} \right)\left( {x + 1} \right)}}
\end{array}\)
BXD:
x -∞ -1 2/3 1 +∞
g(x) - // + 0 - // +
\(\begin{array}{l}
c)h\left( x \right) = 0\\
\to - 8 + 4x - 9x - 3 = 0\\
\to - 5x - 11 = 0\\
\to x = - \dfrac{{11}}{5}
\end{array}\)
BXD:
x -∞ -11/5 +∞
h(x) + 0 -
