Đáp án:
a) \(x \in \left[ {0;1} \right) \cup \left[ {6; + \infty } \right)\)
Giải thích các bước giải:
\(\begin{array}{l}
a)DK:x \ne 1\\
\dfrac{{x\left( {x - 6} \right)}}{{x - 1}} \ge 0
\end{array}\)
BXD:
x -∞ 0 1 6 +∞
f(x) - 0 + // - 0 +
\(KL:x \in \left[ {0;1} \right) \cup \left[ {6; + \infty } \right)\)
\(\begin{array}{l}
b)DK:x \ne \left\{ {2;3} \right\}\\
\dfrac{{5x - 6}}{{{x^2} - 5x + 6}} - 1 < 0\\
\to \dfrac{{5x - 6 - {x^2} + 5x - 6}}{{\left( {x - 2} \right)\left( {x - 3} \right)}} < 0\\
\to \dfrac{{ - {x^2} + 10x - 12}}{{\left( {x - 2} \right)\left( {x - 3} \right)}} < 0
\end{array}\)
BXD:
x -∞ 5-√13 2 3 5+√13 +∞
f(x) - 0 + // - // + 0 -
\(KL:x \in \left( { - \infty ;5 - \sqrt {13} } \right) \cup \left( {2;3} \right) \cup \left( {5 + \sqrt {13} ; + \infty } \right)\)
\(\begin{array}{l}
c)\left\{ \begin{array}{l}
3x > 9\\
x > - 4
\end{array} \right.\\
\to \left\{ \begin{array}{l}
x > 3\\
x > - 4
\end{array} \right.\\
\to x > 3
\end{array}\)
