Đáp án:
d. \(x \in \left( { - \infty ; - 4} \right) \cup \left( {1; + \infty } \right)\)
Giải thích các bước giải:
\(\begin{array}{l}
b.DK:x \ne 2\\
\dfrac{{x + 3 - \left( {x - 1} \right)\left( {x - 2} \right)}}{{x - 2}} \le 0\\
\to \dfrac{{x + 3 - {x^2} + 3x - 2}}{{x - 2}} \le 0\\
\to \dfrac{{ - {x^2} + 4x + 1}}{{x - 2}} \le 0\\
Xét: - {x^2} + 4x + 1 = 0\\
\to \left[ \begin{array}{l}
x = 2 + \sqrt 5 \\
x = 2 - \sqrt 5
\end{array} \right.
\end{array}\)
BXD:
x -∞ \(2 - \sqrt 5 \) 2 \(2 + \sqrt 5 \) +∞
f(x) + 0 - // + 0 -
\(KL:x \in \left[ {2 - \sqrt 5 ;2} \right) \cup \left[ {2 + \sqrt 5 ; + \infty } \right)\)
\(\begin{array}{l}
c.\left\{ \begin{array}{l}
4x \le - 3\\
x \in \left( {\dfrac{{ - 3 - \sqrt {13} }}{2};\dfrac{{ - 3 + \sqrt {13} }}{2}} \right)
\end{array} \right.\\
\to \left\{ \begin{array}{l}
x \le - \dfrac{3}{4}\\
x \in \left( {\dfrac{{ - 3 - \sqrt {13} }}{2};\dfrac{{ - 3 + \sqrt {13} }}{2}} \right)
\end{array} \right.\\
KL:x \in \left( {\dfrac{{ - 3 - \sqrt {13} }}{2}; - \dfrac{3}{4}} \right]\\
d.\left\{ \begin{array}{l}
\left( {x - 1} \right)\left( {x + 4} \right) > 0\\
x\left( {x + 2} \right) > 0
\end{array} \right.\\
\to \left\{ \begin{array}{l}
x \in \left( { - \infty ; - 4} \right) \cup \left( {1; + \infty } \right)\\
x \in \left( { - \infty ; - 2} \right) \cup \left( {0; + \infty } \right)
\end{array} \right.\\
KL:x \in \left( { - \infty ; - 4} \right) \cup \left( {1; + \infty } \right)
\end{array}\)
