Решите уравнение: a) (3x - 2)² - (3x - 4)(3x + 4) = 0; б) 4y² - 81 = 0.

a) $$(3x - 2)^2 - (3x - 4)(3x + 4) = 0$$ $$(9x^2 - 12x + 4) - (9x^2 - 16) = 0$$ $$9x^2 - 12x + 4 - 9x^2 + 16 = 0$$ $$-12x + 20 = 0$$ $$-12x = -20$$ $$x = \frac{-20}{-12} = \frac{5}{3}$$ б) $$4y^2 - 81 = 0$$ $$4y^2 = 81$$ $$y^2 = \frac{81}{4}$$ $$y = \pm \sqrt{\frac{81}{4}} = \pm \frac{9}{2}$$ $$y_1 = \frac{9}{2} = 4.5, y_2 = -\frac{9}{2} = -4.5$$