542. Найдите корни уравнения: a) (2x-3)(5x + 1) = 2x+2/5; б) (Зу - 1)(у + 3) = y(1 + 6y); в) (t - 1)(t + 1) = 2(5t-101/2); г) -z(z + 7) = (z - 2)(z + 2).

a) $$(2x - 3)(5x + 1) = 2x + \frac{2}{5}$$ $$10x^2 + 2x - 15x - 3 = 2x + \frac{2}{5}$$ $$10x^2 - 15x - 3 - \frac{2}{5} = 0$$ $$10x^2 - 15x - \frac{17}{5} = 0$$ $$50x^2 - 75x - 17 = 0$$ $$D = (-75)^2 - 4 \cdot 50 \cdot (-17) = 5625 + 3400 = 9025$$ $$x_1 = \frac{75 + \sqrt{9025}}{2 \cdot 50} = \frac{75 + 95}{100} = \frac{170}{100} = 1.7$$ $$x_2 = \frac{75 - \sqrt{9025}}{2 \cdot 50} = \frac{75 - 95}{100} = \frac{-20}{100} = -0.2$$ Ответ: $$x_1 = 1.7$$, $$x_2 = -0.2$$ б) $$(3y - 1)(y + 3) = y(1 + 6y)$$ $$3y^2 + 9y - y - 3 = y + 6y^2$$ $$3y^2 + 8y - 3 = y + 6y^2$$ $$3y^2 - 7y + 3 = 0$$ $$D = (-7)^2 - 4 \cdot 3 \cdot 3 = 49 - 36 = 13$$ $$y_1 = \frac{7 + \sqrt{13}}{2 \cdot 3} = \frac{7 + \sqrt{13}}{6}$$ $$y_2 = \frac{7 - \sqrt{13}}{2 \cdot 3} = \frac{7 - \sqrt{13}}{6}$$ Ответ: $$y_1 = \frac{7 + \sqrt{13}}{6}$$, $$y_2 = \frac{7 - \sqrt{13}}{6}$$ в) $$(t - 1)(t + 1) = 2(5t - 10 \frac{1}{2})$$ $$t^2 - 1 = 10t - 21$$ $$t^2 - 10t + 20 = 0$$ $$D = (-10)^2 - 4 \cdot 1 \cdot 20 = 100 - 80 = 20$$ $$t_1 = \frac{10 + \sqrt{20}}{2 \cdot 1} = \frac{10 + 2\sqrt{5}}{2} = 5 + \sqrt{5}$$ $$t_2 = \frac{10 - \sqrt{20}}{2 \cdot 1} = \frac{10 - 2\sqrt{5}}{2} = 5 - \sqrt{5}$$ Ответ: $$t_1 = 5 + \sqrt{5}$$, $$t_2 = 5 - \sqrt{5}$$ г) $$-z(z + 7) = (z - 2)(z + 2)$$ $$-z^2 - 7z = z^2 - 4$$ $$2z^2 + 7z - 4 = 0$$ $$D = 7^2 - 4 \cdot 2 \cdot (-4) = 49 + 32 = 81$$ $$z_1 = \frac{-7 + \sqrt{81}}{2 \cdot 2} = \frac{-7 + 9}{4} = \frac{2}{4} = \frac{1}{2}$$ $$z_2 = \frac{-7 - \sqrt{81}}{2 \cdot 2} = \frac{-7 - 9}{4} = \frac{-16}{4} = -4$$ Ответ: $$z_1 = \frac{1}{2}$$, $$z_2 = -4$$