3 Решить уравнение: 1) 3^(x+1) = 27^(x-1) ; 3) 2^(x+3) - 2^(x+1) = 12; 2) 0,2^(x^2+4x-5) = 1; 4) 4⋅2^(2x) - 5⋅2^(x) + 1 = 0.

1) $$3^{x+1} = 27^{x-1}$$ $$3^{x+1} = (3^3)^{x-1}$$ $$3^{x+1} = 3^{3x-3}$$ $$x+1 = 3x - 3$$ $$2x = 4$$ $$x = 2$$ 2) $$0,2^{x^2 + 4x - 5} = 1$$ $$0,2^{x^2 + 4x - 5} = 0,2^0$$ $$x^2 + 4x - 5 = 0$$ $$D = b^2 - 4ac = 4^2 - 4 \cdot 1 \cdot (-5) = 16 + 20 = 36$$ $$x_1 = \frac{-b + \sqrt{D}}{2a} = \frac{-4 + 6}{2} = 1$$ $$x_2 = \frac{-b - \sqrt{D}}{2a} = \frac{-4 - 6}{2} = -5$$ 3) $$2^{x+3} - 2^{x+1} = 12$$ $$2^x \cdot 2^3 - 2^x \cdot 2 = 12$$ $$8 \cdot 2^x - 2 \cdot 2^x = 12$$ $$6 \cdot 2^x = 12$$ $$2^x = 2$$ $$x = 1$$ 4) $$4 \cdot 2^{2x} - 5 \cdot 2^x + 1 = 0$$ $$4 \cdot (2^x)^2 - 5 \cdot 2^x + 1 = 0$$ Замена: $$y = 2^x$$ $$4y^2 - 5y + 1 = 0$$ $$D = b^2 - 4ac = (-5)^2 - 4 \cdot 4 \cdot 1 = 25 - 16 = 9$$ $$y_1 = \frac{-b + \sqrt{D}}{2a} = \frac{5 + 3}{8} = 1$$ $$y_2 = \frac{-b - \sqrt{D}}{2a} = \frac{5 - 3}{8} = \frac{1}{4}$$ $$2^x = 1$$ или $$2^x = \frac{1}{4}$$ $$x = 0$$ или $$x = -2$$ Ответ: 1) x = 2; 2) x = 1 или x = -5; 3) x = 1; 4) x = 0 или x = -2