Решите уравнение:
1) $(3x - 1)(9x^2 + 3x + 1) - 9x(3x^2 - 4) = 17$;
2) $(x + 4)(x^2 - 4x + 16) - x(x - 7)(x + 7) = 15$;
3) $(x + 6)(x^2 - 6x + 36) - x(x - 9)^2 = 4x(4,5x - 13,5)$.
$(3x - 1)(9x^2 + 3x + 1) - 9x(3x^2 - 4) = 17$
$(3x)^3 - 1 - 27x^3 + 36x = 17$
$27x^3 - 27x^3 + 36x = 17 + 1$
36x = 18
$x = \frac{18}{36} = \frac{1}{2}$
$(x + 4)(x^2 - 4x + 16) - x(x - 7)(x + 7) = 15$
$x^3 + 4^3 - x(x^2 - 7^2) = 15$
$x^3 + 64 - x(x^2 - 49) = 15$
$x^3 - x^3 + 49x = 15 - 64$
49x = −49
x = −49 : 49
x = −1
$(x + 6)(x^2 - 6x + 36) - x(x - 9)^2 = 4x(4,5x - 13,5)$
$x^3 + 6^3 - x(x^2 - 18x + 81) = 18x^2 - 54x$
$x^3 + 216 - x^3 + 18x^2 - 81x = 18x^2 - 54x$
$x^3 - x^3 + 18x^2 - 18x^2 + 54x - 81x = -216$
−27x = −216
x = −216 : −27
x = 8
Пожауйста, оцените решение