题目内容
19.| A. | $\left\{\begin{array}{l}{x+1≥0}\\{x-2>0}\end{array}\right.$ | B. | $\left\{\begin{array}{l}{x+1>0}\\{x-2≥0}\end{array}\right.$ | C. | $\left\{\begin{array}{l}{x+1≤0}\\{2-x<0}\end{array}\right.$ | D. | $\left\{\begin{array}{l}{x+1<0}\\{2-x≤0}\end{array}\right.$ |
分析 先根据不等式的解集在数轴上的表示方法求出数轴上表示的解集,再对各选项进行逐一分析即可.
解答 解:∵-1处是空心圆点,且折线向右,
∴x>-1.
∵2处是实心圆点,且折线向右,
∴x>2,
∴数轴上表示的不等式组的解集为$\left\{\begin{array}{l}x>-1\\ x≥2\end{array}\right.$.
A、$\left\{\begin{array}{l}x+1≥0①\\ x-2>0②\end{array}\right.$,由①得,x≥-1,由②得,x>2,不符合题意,故本选项错误;
B、$\left\{\begin{array}{l}x+1>0①\\ x-2≥0②\end{array}\right.$,由①得,x>-1,由②得,x≥2,符合题意,故本选项正确;
C、$\left\{\begin{array}{l}x+1≤0①\\ 2-x<0②\end{array}\right.$,由①得,x≤-1,由②得,x>2,不符合题意,故本选项错误;
D、$\left\{\begin{array}{l}x+1<0①\\ 2-x≤0②\end{array}\right.$,由①得,x<-1,由②得,x≥2,不符合题意,故本选项错误.
故选B.
点评 本题考查的是在数轴上表示不等式的解集,熟知“小于向左,大于向右”是解答此题的关键.
练习册系列答案
相关题目
4.甲乙两个仓库共存粮400吨,现从甲仓库运出存粮的60%,从乙仓库运出存粮的40%.结果乙仓库所余的粮食比甲仓库所余的粮食多20吨.甲乙两个仓库原有存粮各多少吨?设甲仓库存粮x吨,乙仓库存粮y吨,则所列方程组正确的是( )
| A. | $\left\{\begin{array}{l}{x+y=400}\\{60%x-40%y=20}\end{array}\right.$ | B. | $\left\{\begin{array}{l}{x+y=400}\\{40%x-60%y=20}\end{array}\right.$ | ||
| C. | $\left\{\begin{array}{l}{x+y=400}\\{60%y-40%x=20}\end{array}\right.$ | D. | $\left\{\begin{array}{l}{x+y=400}\\{40%-60%x=20}\end{array}\right.$ |