Question

40 ! решите уравнения 1. 2.

Answer 1

$1) \ 3 \cdot 5^{2x - 1} - 2 \cdot 5^{x - 1} = 0,2\\3 \cdot \dfrac{5^{2x}}{5} - 2 \cdot \dfrac{5^{x}}{5} = \dfrac{1}{5} \ \ \ \ \ \ | \cdot 5\\3 \cdot 5^{2x} - 2 \cdot 5^{x} = 1\\3 \cdot 5^{2x} - 2 \cdot 5^{x} - 1 = 0\\5^{x} = t, \ t 0\\3t^{2} - 2t - 1 = 0\\D = (-2)^{2} - 4 \cdot 3 \cdot (-1) = 4 + 12 = 16\\x_{1} = -\dfrac{1}{3} < 0\\ x_{2} = 1$

$5^{x} = 1\\5^{x} = 5^{0}\\x = 0$

ответ: $x = 0$

$2) \ 2^{2x^{2}} + 2^{x^{2} + 2x + 2} = 2^{5 + 4x}\\2^{2x^{2}} + 2^{x^{2} + 2x} \cdot 4 = 32 \cdot 2^{4x} \ \ \ \ \ \ | : 2^{4x}\\2^{2x^{2} - 4x} + 4 \cdot 2^{x^{2} - 2x} = 32\\2^{2(x^{2} - 2x)}+ 4 \cdot 2^{x^{2} - 2x} - 32 = 0\\2^{x^{2} - 2x} = t, \ t 0\\t^{2} + 4t - 32 = 0\\t_{1} = -8 < 0\\t_{2} = 4$

$2^{x^{2} - 2x} = 4\\2^{x^{2} - 2x} = 2^{2}\\x^{2} - 2x = 2\\x^{2} - 2x - 2 = 0\\D = (-2)^{2} - 4 \cdot 1 \cdot (-2) = 4 + 8 = 12\\x_{1} = \dfrac{2 + 2\sqrt{3}}{2} =1 + \sqrt{3} \\x_{2} = \dfrac{2 - 2\sqrt{3}}{2} =1 - \sqrt{3}$

ответ: $x = 1 + \sqrt{3}; \ x = 1 - \sqrt{3}$