Question

(1/2)^(x^2-x) > 8^(x-5)

Answer 1

$(frac{1}{2})^{x^2-x} textgreater 8^{x-5}\2^{x-x^2} textgreater 2^{3x-15}\x-x^2 textgreater 3x-15\x^2+2x-15 textgreater 0\D=4+60=64=8^2\x_1=frac{-2+8}{2}=3\x_2=frac{-2-8}{2}=-5$
Ответ: x∈(–∞; –5)∪(3; +∞)