Решите уравнение:
а) $(x - 6)^2 - x(x + 8) = 2$;
б) $9x(x + 6) - (3x + 1)^2 = 1$;
в) $y(y - 1) - (y - 5)^2 = 2$;
г) $16y(2 - y) + (4y - 5)^2 = 0$.
$(x - 6)^2 - x(x + 8) = 2$
$x^2 - 12x + 36 - x^2 - 8x = 2$
−20x = 2 − 36
−20x = −34
x = 1,7
$9x(x + 6) - (3x + 1)^2 = 1$
$9x^2 + 54x - (9x^2 + 6x + 1) = 1$
$9x^2 + 54x - 9x^2 - 6x - 1 = 1$
48x = 1 + 1
48x = 2
$x = \frac{2}{48} = \frac{1}{24}$
$y(y - 1) - (y - 5)^2 = 2$
$y^2 - y - (y^2 - 10y + 25) = 2$
$y^2 - y - y^2 + 10y - 25 = 2$
9y = 2 + 25
9y = 27
y = 3
$16y(2 - y) + (4y - 5)^2 = 0$
$32y - 16y^2 + 16y^2 - 40y + 25 = 0$
−8y = −25
$y = \frac{25}{8} = 3\frac{1}{8}$
Пожауйста, оцените решение
