1) 6x+(x-5)^2=62+(x+4)(x-12)

Привет! Давай решим уравнения из твоей подготовки к ВПР. Вот пошаговые решения: 1) $6x + (x^2 - 10x + 25) = 62 + (x^2 - 12x + 4x - 48)$ $x^2 - 4x + 25 = x^2 - 8x + 14$ $4x = -11 \Rightarrow x = -2{,}75$ 2) $x + (25x^2 + 20x + 4) = 25 + 25x^2$ $21x = 21 \Rightarrow x = 1$ 3) $(x^2 - 12x + 36) - (x^2 + 8x) = 2$ $-20x = -34 \Rightarrow x = 1{,}7$ 4) $(x^2 - x) - (x^2 - 10x + 25) = 2$ $9x = 27 \Rightarrow x = 3$ 5) $(25x^2 - 9) = 6x + 18x^2 + 13x^2$ $25x^2 - 9 = 31x^2 + 6x$ $6x^2 + 6x + 9 = 0 \Rightarrow x^2 + x + 1{,}5 = 0$. Дискриминант $D = 1 - 6 = -5 < 0$, корней нет. 6) $(8x + 16x^2) - (16x^2 - 9) = 2x$ $8x + 9 = 2x \Rightarrow 6x = -9 \Rightarrow x = -1{,}5$ 7) $(x^2 - 4) - (x^2 - 2x + 1) = x^2 - 5$ $2x - 5 = x^2 - 5 \Rightarrow x^2 - 2x = 0 \Rightarrow x(x - 2) = 0$. Ответ: $x_1 = 0, x_2 = 2$ 8) $(x^2 - 8x + 16) - (x^2 + 16x + 64) = 32$ $-24x - 48 = 32 \Rightarrow -24x = 80 \Rightarrow x = -\frac{80}{24} = -\frac{10}{3} = -3\frac{1}{3}$ 9) $(x^2 - 12x + 36) - (x^2 + 8x + 16) = 24$ $-20x + 20 = 24 \Rightarrow -20x = 4 \Rightarrow x = -0{,}2$ 10) $x^2 + 14x + 49 = x^2 + 7x - 1$ $7x = -50 \Rightarrow x = -\frac{50}{7} \approx -7{,}14$ 11) $16x^2 - (16x^2 - 8x + 1) = 5x - 1$ $8x - 1 = 5x - 1 \Rightarrow 3x = 0 \Rightarrow x = 0$ 12) $(x^2 - 2x + 1) - 7x = 15 + (x^2 - 6x + 9)$ $x^2 - 9x + 1 = x^2 - 6x + 24$ $-3x = 23 \Rightarrow x = -\frac{23}{3} = -7\frac{2}{3}$