Question

(1/4)^2+3x< 8^x-1 -4^2+3x< 2^x-1 2^x -2> ^x-4 120 2^1-3x< 16 25^x-6*5^x +5< 0 решить ? буду : )

Answer 1

Решение: ^ - здесь степень V - корень квадр. Д4.12 log 5 (7-x) = log 5 (3-x) + 1 log 5 (7-x) = log 5 (3-x) + log 5 (5) log 5 (7-x) = log 5 [5*(3-x)] 7-x = 5*(3-x) 7-x = 15 - 5x 5x = 8 x = 5/8 Д4.11 log (x-5) 49 = 2 (x-5)^2 = 49 x^2 - 10x + 25 = 49 x^2 - 10x - 24 = 0 x(1) = 12 x(2) = - 2 Д4.10 2^(3+x) = 0,4 * 5^(3+x) 2^3 * 2^x = 2/5 * 5^3 * 5^x 2^3 * 2^x = 2 * 5^2 * 5^x 2^x /5^x = 2/2^3 * 5^2 (2/5)^x = (5/2)^2 (2/5)^x = (2/5)^(-2) x = -2 Д4.9 (1/3)^(3+x) = 9 [3^(-1)] ^(3+x) = 3^2 3^ (-3-x) = 3^2 -3-x = 2 x = -5 Д4.6 V(6+5x) = x 6+5x = x^2 x^2 - 5x - 6 = 0 x(1) = +6 x(2) = -1 Д4.5 V(1/(5-2x) = 1/3 1/(5-2x) = 1/9 5-2x = 9 2x = -4 x = -2 Д4.4 11x / (2x^2 + 5) = 1 11x = 2x^2 + 5 2x^2 - 11x + 5 = 0 x(1) = +5 x(2) = +1/2 Д4.3 x = (8x+25) / (x+8) x^2 + 8x = 8x + 25 x^2 = 25 x(1) = +5 x(2) = -5 Д4.2 1/7 * x^2 = 9 1/7 1/7 * x^2 = 64/7 x^2 = 64 x(1) = +8 x(2) = -8 Д4.1 (2x+7)^2 = (2x-1)^2 4x^2 + 28x + 49 = 4x^2 - 4x + 1 24x = - 48 x = -2