шите уравнение: (x + 4)² = 3x + 40; (2p – 3)² = 11p – 19; (x + 4)² = 10x + 32; 5y² + 17 = 15(y + 1)²;

Решим уравнение:

a) $$(x + 4)^2 = 3x + 40$$

$$x^2 + 8x + 16 = 3x + 40$$

$$x^2 + 5x - 24 = 0$$

$$D = 5^2 - 4 \cdot 1 \cdot (-24) = 25 + 96 = 121$$

$$x_1 = \frac{-5 + \sqrt{121}}{2 \cdot 1} = \frac{-5 + 11}{2} = \frac{6}{2} = 3$$

$$x_2 = \frac{-5 - \sqrt{121}}{2 \cdot 1} = \frac{-5 - 11}{2} = \frac{-16}{2} = -8$$

Ответ: x₁ = 3; x₂ = -8

б) $$(2p - 3)^2 = 11p - 19$$

$$4p^2 - 12p + 9 = 11p - 19$$

$$4p^2 - 23p + 28 = 0$$

$$D = (-23)^2 - 4 \cdot 4 \cdot 28 = 529 - 448 = 81$$

$$p_1 = \frac{23 + \sqrt{81}}{2 \cdot 4} = \frac{23 + 9}{8} = \frac{32}{8} = 4$$

$$p_2 = \frac{23 - \sqrt{81}}{2 \cdot 4} = \frac{23 - 9}{8} = \frac{14}{8} = \frac{7}{4} = 1.75$$

Ответ: p₁ = 4; p₂ = 1.75

в) $$(x + 4)^2 = 10x + 32$$

$$x^2 + 8x + 16 = 10x + 32$$

$$x^2 - 2x - 16 = 0$$

$$D = (-2)^2 - 4 \cdot 1 \cdot (-16) = 4 + 64 = 68$$

$$x_1 = \frac{2 + \sqrt{68}}{2 \cdot 1} = \frac{2 + 2\sqrt{17}}{2} = 1 + \sqrt{17}$$

$$x_2 = \frac{2 - \sqrt{68}}{2 \cdot 1} = \frac{2 - 2\sqrt{17}}{2} = 1 - \sqrt{17}$$

Ответ: x₁ = 1 + √17; x₂ = 1 - √17

г) $$5y^2 + 17 = 15(y + 1)^2$$

$$5y^2 + 17 = 15(y^2 + 2y + 1)$$

$$5y^2 + 17 = 15y^2 + 30y + 15$$

$$10y^2 + 30y - 2 = 0$$

$$5y^2 + 15y - 1 = 0$$

$$D = 15^2 - 4 \cdot 5 \cdot (-1) = 225 + 20 = 245$$

$$y_1 = \frac{-15 + \sqrt{245}}{2 \cdot 5} = \frac{-15 + 7\sqrt{5}}{10}$$

$$y_2 = \frac{-15 - \sqrt{245}}{2 \cdot 5} = \frac{-15 - 7\sqrt{5}}{10}$$

Ответ: y₁ = (-15 + 7√5)/10; y₂ = (-15 - 7√5)/10