Giải các hệ bất phương trình
a) \(\left\{\begin{matrix} 6x+\frac{5}{7}<4x+7\\ \frac{8x+3}{2}< 2x+5; \end{matrix}\right.\)
b) \(\left\{ {\begin{array}{*{20}{c}}{15x - 2 > 2x + \frac{1}{3}}\\{2(x - 4) < \frac{{3x - 14}}{2}.}\end{array}} \right.\)
Câu a:
\(\left\{ \begin{array}{l}
6x + \frac{5}{7} < 4x + 7\\
\frac{{8x + 3}}{2} < 2x + 5
\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}
2x < 7 - \frac{5}{7}\\
8x + 3 < 4x + 10
\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}
2x < \frac{{44}}{7}\\
4x < 7
\end{array} \right.\)
\(\Leftrightarrow \left\{ \begin{array}{l}
x < \frac{{22}}{7}\\
x < \frac{7}{4}
\end{array} \right. \Leftrightarrow x < \frac{7}{4}\)
Vậy \(S = \left( { - \infty ;\frac{7}{4}} \right)\)
Câu b:
\(\left\{ \begin{array}{l}
15x - 2 > 2x + \frac{1}{3}\\
2(x + 4) < \frac{{3x - 14}}{2}
\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}
13x > \frac{7}{3}\\
4x - 16 < 3x - 14
\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}
x > \frac{7}{{39}}\\
x < 2
\end{array} \right.\)
\( \Leftrightarrow \frac{7}{{39}} < x < 2\)
Vậy \(S = \left( {\frac{7}{{39}};2} \right)\)
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