Đáp án:
c. \(x \in \left( { - 1;\frac{{13 - 3\sqrt {17} }}{2}} \right) \cup \left( {2;8} \right)\)
d. \(\frac{4}{{13}} < x < \frac{{114}}{{50}}\)
Giải thích các bước giải:
\(\begin{array}{l}
a.\left\{ \begin{array}{l}
\frac{{5x + 2 - 3\left( {4 - x} \right)}}{3} \ge 0\\
\frac{{6 - 5x - 13\left( {3x + 1} \right)}}{7} < 0
\end{array} \right.\\
\to \left\{ \begin{array}{l}
5x + 2 - 12 + 3x \ge 0\\
6 - 5x - 39x - 13 < 0
\end{array} \right.\\
\to \left\{ \begin{array}{l}
8x \ge 10\\
44x > - 7
\end{array} \right.\\
\to \left\{ \begin{array}{l}
x \ge \frac{{10}}{8} = \frac{5}{4}\\
x > - \frac{7}{{44}}
\end{array} \right.\\
KL:x \ge \frac{5}{4}\\
b.\left\{ \begin{array}{l}
\frac{{4x - 5 - 7\left( {x + 3} \right)}}{7} < 0\\
\frac{{3x + 8 - 4\left( {2x - 1} \right)}}{4} > 0
\end{array} \right.\\
\to \left\{ \begin{array}{l}
4x - 5 - 7x - 21 < 0\\
3x + 8 - 8x + 4 > 0
\end{array} \right.\\
\to \left\{ \begin{array}{l}
3x > - 26\\
5x < 12
\end{array} \right.\\
\to - \frac{{26}}{3} < x < \frac{{12}}{5}\\
c.DK:x \ne \left\{ { - 1;2} \right\}\\
\left\{ \begin{array}{l}
\frac{{4 - 2x - 8x + 4{x^2} - 3{x^2} - 3x}}{{\left( {x + 1} \right)\left( {2 - x} \right)}} > 0\\
\left( {x - 8} \right)\left( {x + 2} \right) < 0
\end{array} \right.\\
\to \left\{ \begin{array}{l}
\frac{{{x^2} - 13x + 4}}{{\left( {x + 1} \right)\left( {2 - x} \right)}} > 0\left( * \right)\\
x \in \left( { - 2;8} \right)
\end{array} \right.\\
Xét:{x^2} - 13x + 4 = 0\\
\to \left[ \begin{array}{l}
x = \frac{{13 - 3\sqrt {17} }}{2}\\
x = \frac{{13 + 3\sqrt {17} }}{2}
\end{array} \right.
\end{array}\)
BXD:
x -∞ -1 \(\frac{{13 - 3\sqrt {17} }}{2}\) 2 \(\frac{{13 + 3\sqrt {17} }}{2}\) +∞
(*) - // + 0 - // + 0 -
\(\begin{array}{l}
\left( * \right) \to x \in \left( { - 1;\frac{{13 - 3\sqrt {17} }}{2}} \right) \cup \left( {2;\frac{{13 + 3\sqrt {17} }}{2}} \right)\\
KL:x \in \left( { - 1;\frac{{13 - 3\sqrt {17} }}{2}} \right) \cup \left( {2;8} \right)
\end{array}\)
\(\begin{array}{l}
d.\left\{ \begin{array}{l}
\frac{{ - 30x + 9 - 15\left( {2x - 7} \right)}}{{15}} > 0\\
\frac{{2x - 1 - 15x + 5}}{2} < 0
\end{array} \right.\\
\to \left\{ \begin{array}{l}
- 30x + 9 - 20x + 105 > 0\\
- 13x + 4 < 0
\end{array} \right.\\
\to \left\{ \begin{array}{l}
50x < 114\\
13x > 4
\end{array} \right.\\
\to \frac{4}{{13}} < x < \frac{{114}}{{50}}
\end{array}\)
