Решите неравенство 2^2x+1 - 5 * 2^x + 2 >= 0

2 ^ (2 * x + 1) - 5 * 2 ^ x + 2 > = 0;2 ^ (2 * x) * 2 ^ 1 - 5 * 2 ^ x + 2 > = 0;2 * (2 ^ x) ^ 2 - 5 * 2 ^ x + 2 > = 0;2 * (2 ^ x) ^ 2 - 5 * 2 ^ x + 2 = 0;Пусть 2 ^ x = a, тогда:2 * a ^ 2 - 5 * a + 2 = 0;D = b ^ 2 - 4 * a * c = (- 5) ^ 2 - 4 * 2 * 2 = 25 - 16 = 9;a1 = (5 + 3)/(2 * 2) = 8/4 = 2;a2 = (5 - 3)/(2 * 2) = 2/4 = 1/2;Тогда:1) 2 ^ x = a1;2 ^ x = 2;2 ^ x = 2 ^ 1;x = 1;2) 2 ^ x = a2;2 ^ x = 1/2;2 ^ x = 2 ^ (- 1);x = - 1;Отсюда, x < = - 1 и x > = 1.