Решите уравнение:
1) $3x^2 - (2x^2 - 8x) - (x^2 - 3) = x$;
2) $12 - (6 - 9x - x^2) = x^2 + 5x - 14$;
3) $4y^3 - (4y^3 - 8y) - (6y + 3) = 7$;
4) $(y^2 - 4y - 17) - (6y^2 - 3y - 8) = 1 - y - 5y^2$.
$3x^2 - (2x^2 - 8x) - (x^2 - 3) = x$
$3x^2 - 2x^2 + 8x - x^2 + 3 = x$
$(3x^2 - 2x^2 - x^2) + (8x - x) = -3$
$0 + 7x = -3$
$x = -\frac{3}{7}$
$12 - (6 - 9x - x^2) = x^2 + 5x - 14$
$12 - 6 + 9x + x^2 = x^2 + 5x - 14$
$9x + x^2 - x^2 - 5x = -14 - 12 + 6$
9x − 5x = −20
4x = −20
$x = -\frac{20}{4}$
x = −5
$4y^3 - (4y^3 - 8y) - (6y + 3) = 7$
$4y^3 - 4y^3 + 8y - 6y - 3 = 7$
8y − 6y = 7 + 3
2y = 10
y = 10 : 2
y = 5
$(y^2 - 4y - 17) - (6y^2 - 3y - 8) = 1 - y - 5y^2$
$y^2 - 4y - 17 - 6y^2 + 3y + 8 = 1 - y - 5y^2$
$y^2 - 4y - 6y^2 + 3y + y + 5y^2 = 1 - 8 + 17$
$(y^2 - 6y^2 + 5y^2) + (-4y + 3y + y) = 10$
0 ≠ 10, уравнение не имеет корней.
Пожауйста, оцените решение