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Câu hỏi: \(\left\{ \begin{array}{l}\dfrac{{3x + 1}}{2} - \dfrac{{3 - x}}{3} \le \dfrac{{x + 1}}{4} - \dfrac{{2x - 1}}{3}\\3 - \dfrac{{2x + 1}}{5} > x + \dfrac{4}{3}\end{array} \right..\)
Lời giải chi tiết
$\left\{\begin{array}{l}\frac{3 x+1}{2}-\frac{3-x}{3} \leq \frac{x+1}{4}-\frac{2 x-1}{3} \\ 3-\frac{2 x+1}{5}>x+\frac{4}{3}\end{array}\right.$
$\Leftrightarrow\left\{\begin{array}{l}\frac{3}{2} x+\frac{x}{3}-\frac{x}{4}+\frac{2}{3} x \leq \frac{1}{4}+\frac{1}{3}-\frac{1}{2}+1 \\ 3-\frac{1}{5}-\frac{4}{3}>x+\frac{2}{5} x\end{array}\right.$
$\Leftrightarrow\left\{\begin{array}{l}\frac{9}{4} x \leq \frac{13}{12} \\ \frac{22}{15}>\frac{7}{5} x\end{array} \Leftrightarrow\left\{\begin{array}{l}x \leq \frac{13}{27} \\ x<\frac{22}{21}\end{array} \Leftrightarrow x \leq \frac{13}{27}\right.\right.$
$\left\{\begin{array}{l}\frac{3 x+1}{2}-\frac{3-x}{3} \leq \frac{x+1}{4}-\frac{2 x-1}{3} \\ 3-\frac{2 x+1}{5}>x+\frac{4}{3}\end{array}\right.$
$\Leftrightarrow\left\{\begin{array}{l}\frac{3}{2} x+\frac{x}{3}-\frac{x}{4}+\frac{2}{3} x \leq \frac{1}{4}+\frac{1}{3}-\frac{1}{2}+1 \\ 3-\frac{1}{5}-\frac{4}{3}>x+\frac{2}{5} x\end{array}\right.$
$\Leftrightarrow\left\{\begin{array}{l}\frac{9}{4} x \leq \frac{13}{12} \\ \frac{22}{15}>\frac{7}{5} x\end{array} \Leftrightarrow\left\{\begin{array}{l}x \leq \frac{13}{27} \\ x<\frac{22}{21}\end{array} \Leftrightarrow x \leq \frac{13}{27}\right.\right.$