Ответ:
13–5·3x / (9x–12·3x+27) ≥ 0.5⪻⪼t^2-2t+1/t^2-12t+27≤0⪻⪼(t-1)^2/(t-3)(t-9)≤0 t=1 1<x<2
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Ответ:
13–5·3x / (9x–12·3x+27) ≥ 0.5⪻⪼t^2-2t+1/t^2-12t+27≤0⪻⪼(t-1)^2/(t-3)(t-9)≤0 t=1 1<x<2