Tìm tập nghiệm của bpt:
a, (x^2 – 3x – 4)/(3 – 4x) <= 0
b, (x^2 + 3x + 2).(-x + 5) >= 0
c, (x^2 – 5x + 6)/(x^2 + 6x + 9) < 0
Tìm tập nghiệm của bpt:
a, (x^2 – 3x – 4)/(3 – 4x) <= 0
b, (x^2 + 3x + 2).(-x + 5) >= 0
c, (x^2 – 5x + 6)/(x^2 + 6x + 9) < 0
Đáp án:
$\begin{array}{l}
a)\dfrac{{\left( {{x^2} – 3x – 4} \right)}}{{3 – 4x}} \le 0\\
\Rightarrow \left[ \begin{array}{l}
\left\{ \begin{array}{l}
{x^2} – 3x – 4 \ge 0\\
3 – 4x < 0
\end{array} \right.\\
\left\{ \begin{array}{l}
{x^2} – 3x – 4 \le 0\\
3 – 4x > 0
\end{array} \right.
\end{array} \right.\\
\Rightarrow \left[ \begin{array}{l}
\left\{ \begin{array}{l}
\left( {x – 4} \right)\left( {x + 1} \right) \ge 0\\
x > \dfrac{3}{4}
\end{array} \right.\\
\left\{ \begin{array}{l}
\left( {x – 4} \right)\left( {x + 1} \right) \le 0\\
x < \dfrac{3}{4}
\end{array} \right.
\end{array} \right.\\
\Rightarrow \left[ \begin{array}{l}
x \ge 4\\
– 1 \le x < \dfrac{3}{4}
\end{array} \right.\\
b)\left( {{x^2} + 3x + 2} \right)\left( { – x + 5} \right) \ge 0\\
\Rightarrow \left( {x + 1} \right)\left( {x + 2} \right)\left( {x – 5} \right) \le 0\\
\Rightarrow \left[ \begin{array}{l}
x \le – 2\\
– 1 \le x \le 5
\end{array} \right.\\
c)\dfrac{{{x^2} – 5x + 6}}{{{x^2} + 6x + 9}} < 0\\
\Rightarrow \dfrac{{\left( {x – 2} \right)\left( {x – 3} \right)}}{{{{\left( {x + 3} \right)}^2}}} < 0\\
\Rightarrow \left\{ \begin{array}{l}
\left( {x – 2} \right)\left( {x – 3} \right) < 0\\
x \ne – 3
\end{array} \right.\\
\Rightarrow \left\{ \begin{array}{l}
2 < x < 3\\
x \ne – 3
\end{array} \right.\\
\Rightarrow 2 < x < 3
\end{array}$