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If 1 / X-1 / y = 2, find the value of (3x-2xy-3y) / (x-2y-y)
1/x-1/y=(y-x)/xy=2
y-x=2xy
So X-Y = - 2XY
The original formula = [3 (X-Y) - 2XY] / [(X-Y) - 2XY]
=[3(-2xy)-2xy]/[(-2xy)-2xy]
=-8xy/(-4xy)
=2
x. Y and Z are three nonnegative rational numbers, and satisfy 3x + 2Y + Z = 5, x + Y-Z = 2. If 2x + Y-Z = s, what is the value range of S?
We know that x + Y-Z = 2
So Y-Z = 2-x
Because x is a nonnegative rational number
So x ≥ 0
Because 3x + 2Y + Z = 5
So when y = 0, z = 0, X has a maximum of xmax = 5 / 3
be
s = 2x+y-z = 2x+(y-z) = 2x+(2-x) = 2+x ≥2+0 = 2
s = 2+x ≤ 2+5/3 = 11/3
To sum up,
The maximum value of S is Smax = 11 / 3
The minimum value of S is smin = 2
The range of S is [2,11 / 3]
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