252 1) 5^(2x) - 5^(x+5) - 600 = 0; 3) 3^x + 9^(x-1) - 810 = 0; 2) 9^x - 3^(x+6) = 0; 4) 4^x + 2^(2x+1) - 80 = 0.

1) $$5^{2x} - 5^{x+5} - 600 = 0$$ $$5^{2x} - 5^x \cdot 5^5 - 600 = 0$$ Замена: $$y = 5^x$$ $$y^2 - 3125y - 600 = 0$$ $$D = b^2 - 4ac = (-3125)^2 - 4 \cdot 1 \cdot (-600) = 9765625 + 2400 = 9768025$$ $$y_1 = \frac{-b + \sqrt{D}}{2a} = \frac{3125 + \sqrt{9768025}}{2} > 0$$ $$y_2 = \frac{-b - \sqrt{D}}{2a} = \frac{3125 - \sqrt{9768025}}{2} < 0$$ Т.к. $$y = 5^x > 0$$, то $$y_2$$ - не подходит. $$5^x = \frac{3125 + \sqrt{9768025}}{2}$$ $$x = \log_5{\frac{3125 + \sqrt{9768025}}{2}}$$ 2) $$9^x - 3^{x+6} = 0$$ $$9^x = 3^{x+6}$$ $$(3^2)^x = 3^{x+6}$$ $$3^{2x} = 3^{x+6}$$ $$2x = x + 6$$ $$x = 6$$ 3) $$3^x + 9^{x-1} - 810 = 0$$ $$3^x + (3^2)^{x-1} - 810 = 0$$ $$3^x + 3^{2x-2} - 810 = 0$$ $$3^x + \frac{3^{2x}}{3^2} - 810 = 0$$ $$3^x + \frac{(3^x)^2}{9} - 810 = 0$$ Замена: $$y = 3^x$$ $$y + \frac{y^2}{9} - 810 = 0$$ $$9y + y^2 - 7290 = 0$$ $$y^2 + 9y - 7290 = 0$$ $$D = b^2 - 4ac = 9^2 - 4 \cdot 1 \cdot (-7290) = 81 + 29160 = 29241$$ $$y_1 = \frac{-b + \sqrt{D}}{2a} = \frac{-9 + \sqrt{29241}}{2} = \frac{-9 + 171}{2} = \frac{162}{2} = 81 > 0$$ $$y_2 = \frac{-b - \sqrt{D}}{2a} = \frac{-9 - \sqrt{29241}}{2} = \frac{-9 - 171}{2} = \frac{-180}{2} = -90 < 0$$ Т.к. $$y = 3^x > 0$$, то $$y_2$$ - не подходит. $$3^x = 81$$ $$3^x = 3^4$$ $$x = 4$$ 4) $$4^x + 2^{2x+1} - 80 = 0$$ $$(2^2)^x + 2^{2x} \cdot 2 - 80 = 0$$ $$2^{2x} + 2^{2x} \cdot 2 - 80 = 0$$ $$(2^x)^2 + 2 \cdot (2^x)^2 - 80 = 0$$ $$3 \cdot (2^x)^2 - 80 = 0$$ Замена: $$y = 2^x$$ $$3y^2 - 80 = 0$$ $$3y^2 = 80$$ $$y^2 = \frac{80}{3}$$ $$y = \pm \sqrt{\frac{80}{3}}$$ $$y = \pm \sqrt{\frac{16 \cdot 5}{3}}$$ $$y = \pm 4\sqrt{\frac{5}{3}}$$ Т.к. $$y = 2^x > 0$$, то $$y = 4\sqrt{\frac{5}{3}}$$ $$2^x = 4\sqrt{\frac{5}{3}}$$ $$2^x = 2^2\sqrt{\frac{5}{3}}$$ $$x = \log_2{(4\sqrt{\frac{5}{3}})}$$ Ответ: 1) $$x = \log_5{\frac{3125 + \sqrt{9768025}}{2}}$$; 2) x = 6; 3) x = 4; 4) $$x = \log_2{(4\sqrt{\frac{5}{3}})}$$