Ответ:
Решить систему неравенств .
[tex]\left\{\begin{array}{l}\bf \dfrac{x}{3}-\dfrac{x}{4} < 1\ ,\\\ \bf 2x-\dfrac{x}{2}\geq 10\end{array}\right[/tex]
Умножим первое неравенство на 12, а второе на 2, получим[tex]\left\{\begin{array}{l}\bf 4x-3x < 12\\\bf 4x-x\geq 20\end{array}\right\ \ \left\{\begin{array}{l}\bf x < 12\\\bf 3x\geq 20\end{array}\right\ \ \left\{\begin{array}{l}\bf x < 12\\\bf x\geq 6\dfrac{2}{3}\end{array}\right\ \ \ \Rightarrow \\\\\\\boldsymbol{x\in [\ 6\dfrac{2}{3}\ ;\ 12\ )}[/tex]
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Ответ:
Решить систему неравенств .
[tex]\left\{\begin{array}{l}\bf \dfrac{x}{3}-\dfrac{x}{4} < 1\ ,\\\ \bf 2x-\dfrac{x}{2}\geq 10\end{array}\right[/tex]
Умножим первое неравенство на 12, а второе на 2, получим[tex]\left\{\begin{array}{l}\bf 4x-3x < 12\\\bf 4x-x\geq 20\end{array}\right\ \ \left\{\begin{array}{l}\bf x < 12\\\bf 3x\geq 20\end{array}\right\ \ \left\{\begin{array}{l}\bf x < 12\\\bf x\geq 6\dfrac{2}{3}\end{array}\right\ \ \ \Rightarrow \\\\\\\boldsymbol{x\in [\ 6\dfrac{2}{3}\ ;\ 12\ )}[/tex]