3. Решите уравнение: a) (2x-5)² = 4x²; 6) (3x + 1)² - 3x(3x + 1); B) (7x-5)²=7(x + 1)².

Решим каждое уравнение:

1. a) $$(2x-5)^2 = 4x^2$$
   $$4x^2 - 20x + 25 = 4x^2$$
   $$4x^2 - 4x^2 - 20x = -25$$
   $$-20x = -25$$
   $$x = \frac{-25}{-20} = \frac{5}{4} = 1,25$$
2. б) $$(3x + 1)^2 - 3x(3x + 1) = 0$$
   $$9x^2 + 6x + 1 - 9x^2 - 3x = 0$$
   $$9x^2 - 9x^2 + 6x - 3x = -1$$
   $$3x = -1$$
   $$x = -\frac{1}{3}$$
3. в) $$(7x-5)^2=7(x + 1)^2$$
   $$49x^2 - 70x + 25 = 7(x^2 + 2x + 1)$$
   $$49x^2 - 70x + 25 = 7x^2 + 14x + 7$$
   $$49x^2 - 7x^2 - 70x - 14x = 7 - 25$$
   $$42x^2 - 84x = -18$$
   $$42x^2 - 84x + 18 = 0$$
   $$21x^2 - 42x + 9 = 0$$
   $$D = (-42)^2 - 4 \cdot 21 \cdot 9 = 1764 - 756 = 1008$$
   $$x_1 = \frac{42 + \sqrt{1008}}{2 \cdot 21} = \frac{42 + 12\sqrt{7}}{42} = 1 + \frac{2\sqrt{7}}{7}$$
   $$x_2 = \frac{42 - \sqrt{1008}}{2 \cdot 21} = \frac{42 - 12\sqrt{7}}{42} = 1 - \frac{2\sqrt{7}}{7}$$

Ответ:

<ol>
<li>$$x = 1,25$$</li>
<li>$$x = -\frac{1}{3}$$</li>
<li>$$x_1 = 1 + \frac{2\sqrt{7}}{7}; x_2 = 1 - \frac{2\sqrt{7}}{7}$$</li>
</ol>