Question

4．不等式組$\left\{\begin{array}{l}{3（x+1）＞x-1}\\{-\frac{2}{3}x+3≥2}\end{array}\right.$的整數(shù)解的和是（　�。�

• A． -1
• B． 1
• C． 0
• D． 1

Answer 1

分析 先解出不等式組的解集，從而可以得到不等式組的整數(shù)解，從而可以得到不等式組$\left\{\begin{array}{l}{3（x+1）＞x-1}\\{-\frac{2}{3}x+3≥2}\end{array}\right.$的整數(shù)解的和．

解答 解：$\left\{\begin{array}{l}{3（x+1）＞x-1}\\{-\frac{2}{3}x+3≥2}\end{array}\right.$
解得，-2＜x≤$\frac{3}{2}$，
∴$\left\{\begin{array}{l}{3（x+1）＞x-1}\\{-\frac{2}{3}x+3≥2}\end{array}\right.$的整數(shù)解是x=-1，x=0，x=1，
∵（-1）+0+1=0，
故$\left\{\begin{array}{l}{3（x+1）＞x-1}\\{-\frac{2}{3}x+3≥2}\end{array}\right.$的整數(shù)解得和是0，
故選C．

點(diǎn)評 本題考查一元一次不等式組的整數(shù)解，解題的關(guān)鍵是明確解一元一次不等式組的方法．