Bài tập 4.28 trang 109 SBT Toán 10

\(\begin{array}{l}
\left\{ \begin{array}{l}
\frac{{3x + 1}}{2} - \frac{{3 - x}}{3} \le \frac{{x + 1}}{4} - \frac{{2x - 1}}{3}\\
3 - \frac{{2x + 1}}{5} > x + \frac{4}{3}
\end{array} \right.\\
 \Leftrightarrow \left\{ \begin{array}{l}
\frac{3}{2}x + \frac{x}{3} - \frac{x}{4} + \frac{2}{3}x \le \frac{1}{4} + \frac{1}{3} - \frac{1}{2} + 1\\
3 - \frac{1}{5} - \frac{4}{3} > x + \frac{2}{5}x
\end{array} \right.\\
 \Leftrightarrow \left\{ \begin{array}{l}
\frac{9}{4}x \le \frac{{13}}{{12}}\\
\frac{{22}}{{15}} > \frac{7}{5}x
\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}
x \le \frac{{13}}{{27}}\\
x < \frac{{22}}{{11}}
\end{array} \right. \Leftrightarrow x \le \frac{{13}}{{27}}
\end{array}\)

-- Mod Toán 10 HỌC247