Log(x)17+log(17)x<=2 Решить неравенство

log17(x) + logx(17) <= 2x > 0, x != 1logx(17) = 1log17(x)log17(x) + 1log17(x) <= 2 log17(x) = a a + 1a <= 2a + 1a - 2 <= 0a^2 + 1 - 2a <= 0a1 + a2 = 2a1*a2 = 1a = 1- 1 + a <= 1log17(x) <= 1x <= 17 c учетом x > 0, x != 1 получаем ответ :0 < x <= 17, x != 1