Đáp án:
c) \(x \in \left( { - \infty ; - 1} \right) \cup \left( {2;3} \right)\)
Giải thích các bước giải:
\(c)DK:x \ne - 1\)
BXD:
x -∞ -1 2 3 +∞
3-x + / + / + 0 -
x-2 - / - / + / +
x+1 - 0 + / + / +
f(x) + // - 0 + 0 -
\(KL:x \in \left( { - \infty ; - 1} \right) \cup \left( {2;3} \right)\)
\(\begin{array}{l}
d)DK:x \ne \left\{ {\dfrac{1}{2};2} \right\}\\
\dfrac{2}{{x - 2}} < \dfrac{5}{{2x - 1}}\\
\to \dfrac{{4x - 2 - 5x + 10}}{{\left( {x - 2} \right)\left( {2x - 1} \right)}} < 0\\
\to \dfrac{{8 - x}}{{\left( {x - 2} \right)\left( {2x - 1} \right)}} < 0
\end{array}\)
BXD:
x -∞ 1/2 2 8 +∞
8-x + / + / + 0 -
x-2 - / - 0 + / +
2x-1 - 0 + / + / +
f(x) + // - // + 0 -
\(KL:x \in \left( {\dfrac{1}{2};2} \right) \cup \left( {8; + \infty } \right)\)
