3) x² + 2x = 4-1 x2 + 2x = 3 X212X-3=0 D/3 1)√y=- x (x=25 2) (4x²-xy = 26 23x2+xy=2 3) Jy-x=5

Решаем уравнения.

3) $$x^2+2x=4-1$$

$$x^2+2x=3$$

$$x^2+2x-3=0$$

$$D = b^2 - 4ac = 2^2 - 4 \cdot 1 \cdot (-3) = 4 + 12 = 16$$

$$x_1 = \frac{-b + \sqrt{D}}{2a} = \frac{-2 + \sqrt{16}}{2 \cdot 1} = \frac{-2 + 4}{2} = \frac{2}{2} = 1$$

$$x_2 = \frac{-b - \sqrt{D}}{2a} = \frac{-2 - \sqrt{16}}{2 \cdot 1} = \frac{-2 - 4}{2} = \frac{-6}{2} = -3$$

Ответ: $$x_1 = 1, x_2 = -3$$