Đáp án:
f) \(x \in \left( { - \infty ; - 3} \right] \cup \left( { - 1;2} \right]\)
Giải thích các bước giải:
a) BXD:
x -∞ -3 1/2 +∞
f(x) + 0 - 0 +
\(KL:x \in \left[ { - 3;\dfrac{1}{2}} \right]\)
b) BXD:
x -∞ -3 -2 -1 +∞
f(x) + 0 - 0 + 0 -
\(KL:x \in \left( { - \infty ; - 3} \right) \cup \left( { - 2; - 1} \right)\)
c) BXD:
x -∞ -1 2 3 +∞
f(x) + 0 - 0 + 0 -
\(KL:x \in \left( { - 1;2} \right) \cup \left( {3; + \infty } \right)\)
\(\begin{array}{l}
d)DK:x \ne 2\\
\dfrac{{4x - 3}}{{x - 2}} - 5 > 0\\
\to \dfrac{{4x - 3 - 5x + 10}}{{x - 2}} > 0\\
\to \dfrac{{ - x + 7}}{{x - 2}} > 0
\end{array}\)
BXD:
x -∞ 2 7 +∞
f(x) - // + 0 -
\(KL:x \in \left( {2;7} \right)\)
\(e)DK:x \ne - 1\)
BXD:
x -∞ -3 -1 2 +∞
f(x) + 0 - // + 0 -
\(KL:x \in \left( { - \infty ; - 3} \right] \cup \left( { - 1;2} \right]\)
