题目内容
16.不等式组$\left\{\begin{array}{l}{x+y≥1}\\{x-2y≤4}\end{array}\right.$的解集为D,下列命题中正确的是( )| A. | ?(x,y)∈D,x+2y≤-1 | B. | ?(x,y)∈D,x+2y≥-2 | C. | ?(x,y)∈D,x+2y≤3 | D. | ?(x,y)∈D,x+2y≥2 |
分析 化简不等式组$\left\{\begin{array}{l}{x+y≥1}\\{x-2y≤4}\end{array}\right.$,即可得出正确的结论
解答 解:∵不等式组 $\left\{\begin{array}{l}{x+y≥1…①}\\{x-2y≤4…②}\end{array}\right.$,
∴$\left\{\begin{array}{l}{x+y≥1…①}\\{-x+2y≥-4…③}\end{array}\right.$,
∴$\left\{\begin{array}{l}{x+y≥1…①}\\{y≥-1…*}\end{array}\right.$,
∴x+2y≥0;
即x+2y≥-2.
∴若$\left\{\begin{array}{l}{x+y≥1}\\{x-2y≤4}\end{array}\right.$的解集为D时,?(x,y)∈D,x+2y≥-2成立.
故选:B.
点评 本题考查了不等式组的解法与应用问题,也考查了全称命题与特称命题的应用问题,是基础题目.
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