Question

15．不等式組$\left\{\begin{array}{l}2x-1＞4\\ x-6≤-2\end{array}\right.$的整數(shù)解是3和4．

Answer 1

分析 解不等式組得出x的取值范圍，再取其內(nèi)的整數(shù)即可得出結(jié)論．

解答 解：$\left\{\begin{array}{l}{2x-1＞4①}\\{x-6≤-2②}\end{array}\right.$．
解不等式①得：x＞$\frac{5}{2}$；
解不等式②得：x≤4．
∴不等式組的解為$\frac{5}{2}$＜x≤4，
∴不等式組$\left\{\begin{array}{l}2x-1＞4\\ x-6≤-2\end{array}\right.$的整數(shù)解是3和4．

點(diǎn)評(píng) 本題考查了一元一次不等式組的整數(shù)解，熟練掌握一元一次不等式組的解法是解題的關(guān)鍵．