当M取什么值时,使方程组 X+2Y=1-M/2 3X+4Y=2M 的解使X-Y小于0成立
人气：378 ℃ 时间：2019-08-20 20:41:20
解答
X+2Y=1-M/2 .(1)
2X+4Y=1-M
3X+4Y=2M .(2)
X=2M-1+M=3M-1
Y=[1-M-2(3M-1)]/4=(3-7M)/4
X-Y=3M-1-(3-7M)/4=(19M-7)/4<0
19M-7<0
M<7/19
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