Z=3ˣ A=(13-5z)/(z²-12z+27)-0.5(z²-12z+27)≥0
[13-5z-0.5(z²-12z+27)]/(z²-12z+27)≥0
[13-5z-0.5z²+6z-13.5]=[-0.5z²+z-0.5]
(z²-2z+1)/(z²-12z+27)≤0 z²-12z+27=0 z1=9 z2=3
(z-1)²/(z-3)(z-9)≤0 z=1
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+ - +
z∈{1}∪(3;9)
z=3ˣ z=1 x=0 z=3 x=1 z=9 x=2
x∈{0}∪(1;2)