x,y,z为三个非负有理数,且满足3x+2y+z=5,x+y-z=2,若2x+y-z=s,则s的取值范围是什么?

已知 x+y-z = 2
所以 y-z = 2-x
因为 x是非负有理数
所以 x ≥0
又因为 3x＋2y＋z＝5
所以当y = 0 ,z = 0时 x有最大值为 xmax = 5/3
则
s = 2x+y-z = 2x+(y-z) = 2x+(2-x) = 2+x ≥2+0 = 2
s = 2+x ≤ 2+5/3 = 11/3
综上,
s的最大值为 smax = 11/3
s的最小值为 smin = 2
s的取值范围是[2,11/3]