5. Решите уравнение: a) $(2x - 5)^2 - (2x - 3)(2x + 3) = 0$ b) $9y^2 - 25 = 0$

a) $(2x - 5)^2 - (2x - 3)(2x + 3) = 0$ $(4x^2 - 20x + 25) - (4x^2 - 9) = 0$ $4x^2 - 20x + 25 - 4x^2 + 9 = 0$ $-20x + 34 = 0$ $-20x = -34$ $x = \frac{-34}{-20} = \frac{17}{10} = 1.7$ b) $9y^2 - 25 = 0$ $9y^2 = 25$ $y^2 = \frac{25}{9}$ $y = \pm \sqrt{\frac{25}{9}} = \pm \frac{5}{3}$ **Ответ:** a) $x = 1.7$ b) $y = \frac{5}{3}$ или $y = -\frac{5}{3}$