Question

Можно решение? поочередно все действия) заранее **

Answer 1

$a) 4^{x+1} + 4^{1-x} - 10 = 0\\\\4^{1+x} + 4^{1-x} = 10\\\\4*4^x + \frac{4}{4^x} = 10\\\\4^x = t\\\\4t + \frac{4}{t} = 10\\\\4t^2 + 4 - 10t = 0\\\\D = 100 - 64 = 36\\\\\sqrt{D} = 6\\\\t_1 = \frac{10-6}{8} = \frac{4}{8} = \frac{1}{2}\\\\t_2 = \frac{10+6}{8} = \frac{16}{8} = 2\\\\4^x = \frac{1}{2}\\\\4^x = 4^{-\frac{1}{2}}\\\\x_1 = -0,5\\\\4^x = 2\\\\4^x = 4^{\frac{1}{2}}\\\\x_2 = 0,5\\\\Otvet: \boxed {\pm\frac{1}{2}}$

$3^{x+1} + 3^{1-x} - 10 = 0\\\\3*3^x + \frac{3}{3^x} - 10 = 0\\\\3^x = t\\\\3t + \frac{3}{t} - 10 = 0\\\\3t^2 + 3 - 10t = 0\\\\ D = 100 - 36 = 64\\\\t_1 = \frac{10-8}{6} = \frac{2}{6} = \frac{1}{3}\\\\t_2 = \frac{10+8}{6} = \frac{18}{6} = 3\\\\3^x = \frac{1}{3}\\\\3^x = 3^{-1}\\\\ x = -1\\\\3^x = 3\\\\3^x = 3^1\\\\x = 1\\\\Otvet: \boxed {\pm 1}$

$c) 2^{2+x} - 2^{2-x} = 15\\\\4*2^x - \frac{4}{2^x} -15 = 0\\\\2^x = t\\\\4t-\frac{4}{t} - 15 = 0\\\\4t^2 - 4 - 15t = 0\\\\D = 225 + 64 = 289\\\\\sqrt{D} = 17\\\\t_1 = \frac{15-17}{8} = -\frac{2}{8} = -\frac{1}{4}\\\\t_2 = \frac{15+17}{8} = \frac{32}{8} = 4\\\\2^x = -\frac{1}{4}\\\\2^x = 4\\\\2^x = 2^2\\\\x = 2\\\\Otvet: \boxed{2}$

Отрицательное значение t не подходит, так ни одного рационального числа при котором 2 в некоторой степени даст -0,25 не существует. Возможен сложный иррациональный корень, но вряд ли это из школьной программы.

$d) 3^{2-x} + 3^{x+1} = 12\\\\\frac{9}{3^x} + 3*3^x - 12 = 0\\\\3^x = t\\\\\frac{9}{t} + 3t - 12 = 0\\\\9 + 3t^2 - 12t = 0\\\\D = 144 - 108 = 36\\\\\sqrt{D} = 6\\\\t_1 = \frac{12-6}{6} = \frac{6}{6} = 1\\\\t_2 = \frac{12+6}{6} = \frac{18}{6} = 3\\\\3^x = 1\\\\3^x = 3^0\\\\x = 0\\\\3^x = 3\\\\3^x = 3^1\\\\x = 1\\\\Otvet: \boxed{x_1 = 0; x_2 = 1}$