543. Решите уравнение: д) (x + 1)² = 7918 – 2x; е) (m + 2)² = 3131 – 2m; ж) (x + 1)² = (2x – 1)²; з) (n – 2)² + 48 = (2– 3n)².

д) $$(x + 1)^2 = 7918 - 2x$$

$$x^2 + 2x + 1 = 7918 - 2x$$

$$x^2 + 4x - 7917 = 0$$

$$D = 4^2 - 4 \cdot 1 \cdot (-7917) = 16 + 31668 = 31684$$

$$x_1 = \frac{-4 + \sqrt{31684}}{2 \cdot 1} = \frac{-4 + 178}{2} = \frac{174}{2} = 87$$

$$x_2 = \frac{-4 - \sqrt{31684}}{2 \cdot 1} = \frac{-4 - 178}{2} = \frac{-182}{2} = -91$$

Ответ: $$x_1 = 87$$, $$x_2 = -91$$

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е) $$(m + 2)^2 = 3131 - 2m$$

$$m^2 + 4m + 4 = 3131 - 2m$$

$$m^2 + 6m - 3127 = 0$$

$$D = 6^2 - 4 \cdot 1 \cdot (-3127) = 36 + 12508 = 12544$$

$$m_1 = \frac{-6 + \sqrt{12544}}{2 \cdot 1} = \frac{-6 + 112}{2} = \frac{106}{2} = 53$$

$$m_2 = \frac{-6 - \sqrt{12544}}{2 \cdot 1} = \frac{-6 - 112}{2} = \frac{-118}{2} = -59$$

Ответ: $$m_1 = 53$$, $$m_2 = -59$$

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ж) $$(x + 1)^2 = (2x - 1)^2$$

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

$$3x^2 - 6x = 0$$

$$3x(x - 2) = 0$$

$$x_1 = 0$$, $$x - 2 = 0$$

$$x_2 = 2$$

Ответ: $$x_1 = 0$$, $$x_2 = 2$$

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з) $$(n - 2)^2 + 48 = (2 - 3n)^2$$

$$n^2 - 4n + 4 + 48 = 4 - 12n + 9n^2$$

$$8n^2 - 8n - 48 = 0$$

$$n^2 - n - 6 = 0$$

$$D = (-1)^2 - 4 \cdot 1 \cdot (-6) = 1 + 24 = 25$$

$$n_1 = \frac{1 + \sqrt{25}}{2 \cdot 1} = \frac{1 + 5}{2} = \frac{6}{2} = 3$$

$$n_2 = \frac{1 - \sqrt{25}}{2 \cdot 1} = \frac{1 - 5}{2} = \frac{-4}{2} = -2$$

Ответ: $$n_1 = 3$$, $$n_2 = -2$$