Đáp án:
6) \(x \in \left[ { - 5; - 2} \right] \cup \left[ {2; + \infty } \right)\)
Giải thích các bước giải:
1) Xét:
\(\begin{array}{l}
\left( {x + 1} \right)\left( {x - 2} \right) = 0\\
\to \left[ \begin{array}{l}
x = - 1\\
x = 2
\end{array} \right.
\end{array}\)
BXD:
x -∞ -1 2 +∞
f(x) + 0 - 0 +
\(KL:x \in \left( { - \infty ; - 1} \right] \cup \left[ {2; + \infty } \right)\)
2) Xét:
\(\begin{array}{l}
\left( {2x - 7} \right)\left( {4 - 5x} \right) = 0\\
\to \left[ \begin{array}{l}
x = \dfrac{7}{2}\\
x = \dfrac{4}{5}
\end{array} \right.
\end{array}\)
BXD:
x -∞ 4/5 7/2 +∞
f(x) - 0 + 0 -
\(KL:x \in \left( { - \infty ;\dfrac{4}{5}} \right] \cup \left[ {\dfrac{7}{2}; + \infty } \right)\)
3) BXD:
x -∞ -7/2 0 3 +∞
f(x) - 0 + 0 - 0 +
\(KL:x \in \left( { - \dfrac{7}{2};0} \right) \cup \left( {3; + \infty } \right)\)
4) BXD:
x -∞ -1/2 3 +∞
f(x) - 0 + 0 -
\(KL:x \in \left( {\dfrac{1}{2};3} \right)\)
5) BXD:
x -∞ 0 1/2 3 +∞
f(x) - 0 + 0 - 0 +
\(KL:x \in \left( { - \infty ;0} \right] \cup \left[ {\dfrac{1}{2};3} \right]\)
6) BXD:
x -∞ -5 -2 2 +∞
f(x) - 0 + 0 - 0 +
\(KL:x \in \left[ { - 5; - 2} \right] \cup \left[ {2; + \infty } \right)\)
