ВАРИАНТ 1 1) x² + 5x = 0 2) x²-4=0 3) 3x + 2x²- 5 = 0 4) x² + 2 + 3x = 0 5) x² + 4x + 4 = 0 6) 3x² + 8x = 3 7) 6a² + 2 = 6a

Предмет: Математика 1) $$x^2 + 5x = 0$$ $$x(x+5) = 0$$ $$x_1 = 0$$ $$x+5 = 0$$ $$x_2 = -5$$ 2) $$x^2 - 4 = 0$$ $$x^2 = 4$$ $$x_1 = 2$$ $$x_2 = -2$$ 3) $$3x + 2x^2 - 5 = 0$$ $$2x^2 + 3x - 5 = 0$$ $$D = b^2 - 4ac = 3^2 - 4 \cdot 2 \cdot (-5) = 9 + 40 = 49$$ $$x_1 = \frac{-b + \sqrt{D}}{2a} = \frac{-3 + \sqrt{49}}{2 \cdot 2} = \frac{-3 + 7}{4} = \frac{4}{4} = 1$$ $$x_2 = \frac{-b - \sqrt{D}}{2a} = \frac{-3 - \sqrt{49}}{2 \cdot 2} = \frac{-3 - 7}{4} = \frac{-10}{4} = -2.5$$ 4) $$x^2 + 2 + 3x = 0$$ $$x^2 + 3x + 2 = 0$$ $$D = b^2 - 4ac = 3^2 - 4 \cdot 1 \cdot 2 = 9 - 8 = 1$$ $$x_1 = \frac{-b + \sqrt{D}}{2a} = \frac{-3 + \sqrt{1}}{2 \cdot 1} = \frac{-3 + 1}{2} = \frac{-2}{2} = -1$$ $$x_2 = \frac{-b - \sqrt{D}}{2a} = \frac{-3 - \sqrt{1}}{2 \cdot 1} = \frac{-3 - 1}{2} = \frac{-4}{2} = -2$$ 5) $$x^2 + 4x + 4 = 0$$ $$(x+2)^2 = 0$$ $$x+2 = 0$$ $$x = -2$$ 6) $$3x^2 + 8x = 3$$ $$3x^2 + 8x - 3 = 0$$ $$D = b^2 - 4ac = 8^2 - 4 \cdot 3 \cdot (-3) = 64 + 36 = 100$$ $$x_1 = \frac{-b + \sqrt{D}}{2a} = \frac{-8 + \sqrt{100}}{2 \cdot 3} = \frac{-8 + 10}{6} = \frac{2}{6} = \frac{1}{3}$$ $$x_2 = \frac{-b - \sqrt{D}}{2a} = \frac{-8 - \sqrt{100}}{2 \cdot 3} = \frac{-8 - 10}{6} = \frac{-18}{6} = -3$$ 7) $$6a^2 + 2 = 6a$$ $$6a^2 - 6a + 2 = 0$$ $$3a^2 - 3a + 1 = 0$$ $$D = b^2 - 4ac = (-3)^2 - 4 \cdot 3 \cdot 1 = 9 - 12 = -3$$ Так как дискриминант меньше нуля, то уравнение не имеет действительных решений.