实数x,y满足|x+y|≤1,|x-y|≤1.证明|x|+|y|≤1.
已知实数x,y满足y≤1y≥|x−1|
设实数x,y满足不等式组y+x≤1y-x≤1y≥0
若实数x,y满足y≥1y≤2x−1x+y≤m.
实数x,y满足x
若实数x,y满足不等式组x-y+1≥0x+y-1≤0y≥0
若实数x、y,满足x≥0y≥04x+3y≤12
已知实数x,y满足约束条件x−y≤12x+y≤4x≥1
若实数x,y满足x≥1x+y≤4x−2y+c≤0
若实数x,y满足x−y+1≥0x+y≥0x≤0
若实数x,y满足约束条件x-y≥0,x+y≥0,2x+y≤1,则y/(x+1)的最大值是多少,
实数x,y满足:|x+y|
已知实数x,y满足条件x≥0,y≥x,3x+4y≤12,则(x+2y+3)/(x +1)的最大值是