Ответы на вопрос:
4^(x-3) - 71*2^(x-6) + 7 ≤ 0 2^(2x-6) - 71*2^(x-6) + 7 ≤ 0 (1/64)*2^(2x) - (71/64)*2^(x) + 7 ≤ 0 замена: 2^(x) = t > 0 (1/64)*t^2 - (71/64)*t + 7 ≤ 0 t^2 - 71t + 448 ≤ 0 d = 3249 = 57^2 t1 = (71-57)/2 = 7 > 0 t2 = (71+57)/2 = 64 > 0 7 ≤ t ≤ 64 7 ≤ 2^x ≤ 64 log2(7) ≤ x ≤ 6
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