Задание 5. Найдите корень уравнения. 1) x²+x+6=-x²-3x+(-2+2x²); 2) -3x²+5x-3=-x2+3x+(2-2x²); 3) 3x²-4x+7=x²-5x+(-1+2x²); 4) -4x²+2x+6=-2x2+3x-(-3+2x²).

Решим уравнения:

1. $$x^2+x+6=-x^2-3x+(-2+2x^2)$$ $$x^2+x+6=-x^2-3x-2+2x^2$$ $$x^2+x+6=-x^2-3x-2+2x^2$$ $$x^2+x+6 - 2x^2 +x^2 +3x +2 = 0$$ $$4x + 8 = 0$$ $$4x = -8$$ $$x = -2$$
2. $$-3x^2+5x-3=-x^2+3x+(2-2x^2)$$ $$-3x^2+5x-3=-x^2+3x+2-2x^2$$ $$-3x^2+5x-3 +2x^2 + x^2 -3x -2 = 0$$ $$2x - 5 = 0$$ $$2x = 5$$ $$x = 2.5$$
3. $$3x^2-4x+7=x^2-5x+(-1+2x^2)$$ $$3x^2-4x+7=x^2-5x-1+2x^2$$ $$3x^2-4x+7 -2x^2 -x^2 +5x +1 = 0$$ $$x + 8 = 0$$ $$x = -8$$
4. $$-4x^2+2x+6=-2x^2+3x-(-3+2x^2)$$ $$-4x^2+2x+6=-2x^2+3x+3-2x^2$$ $$-4x^2+2x+6 + 4x^2 -3x -3 = 0$$ $$-x + 3 = 0$$ $$x = 3$$

Ответ: 1) x = -2; 2) x = 2.5; 3) x = -8; 4) x = 3