Решите уравнение:
1) (3x − 1)(4x + 5) − (2x + 3)(6x + 1) = 4;
2) $8x(2x + 7) - (4x + 3)^2 = 15$.
(3x − 1)(4x + 5) − (2x + 3)(6x + 1) = 4
$12x^2 - 4x + 15x - 5 - (12x^2 + 18x + 2x + 3) = 4$
$12x^2 + 11x - 5 - 12x^2 - 18x - 2x - 3 = 4$
−9x − 8 = 4
−9x = 4 + 8
−9x = 12
$x = -\frac{12}{9} = -\frac{4}{3} = -1\frac{1}{3}$
$8x(2x + 7) - (4x + 3)^2 = 15$
$16x^2 + 56x - (16x^2 + 24x + 9) = 15$
$16x^2 + 56x - 16x^2 - 24x - 9 = 15$
32x = 15 + 9
32x = 24
$x = \frac{24}{32} = \frac{3}{4}$
Пожауйста, оцените решение