543. Решите уравнение: a) (x + 4)² = 3x + 40; б) (2p – 3)² = 11p – 19; в) 3(x + 4)² = 10x + 32; г) 15у² + 17 = 15(y + 1)²;

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: $$x_1 = 3$$, $$x_2 = -8$$

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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: $$p_1 = 4$$, $$p_2 = 1.75$$

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в) $$3(x + 4)^2 = 10x + 32$$

$$3(x^2 + 8x + 16) = 10x + 32$$

$$3x^2 + 24x + 48 = 10x + 32$$

$$3x^2 + 14x + 16 = 0$$

$$D = 14^2 - 4 \cdot 3 \cdot 16 = 196 - 192 = 4$$

$$x_1 = \frac{-14 + \sqrt{4}}{2 \cdot 3} = \frac{-14 + 2}{6} = \frac{-12}{6} = -2$$

$$x_2 = \frac{-14 - \sqrt{4}}{2 \cdot 3} = \frac{-14 - 2}{6} = \frac{-16}{6} = -\frac{8}{3}$$

Ответ: $$x_1 = -2$$, $$x_2 = -\frac{8}{3}$$

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г) $$15y^2 + 17 = 15(y + 1)^2$$

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

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

$$17 = 30y + 15$$

$$30y = 2$$

$$y = \frac{2}{30} = \frac{1}{15}$$

Ответ: $$y = \frac{1}{15}$$