Решите уравнение:
а) x(x − 1) − x(x − 3) = 12;
б) $(x + 1)(x + 2) - x^2 = 5x + 4$;
в) $(x - 4)^2 = x^2 - 16$;
г) $(x + 1)^2 = x^2 + 1$.
x(x − 1) − x(x − 3) = 12
$x^2 - x - x^2 + 3x = 12$
2x = 12
x = 6
$(x + 1)(x + 2) - x^2 = 5x + 4$
$x^2 + x + 2x + 2 - x^2 - 5x = 4$
−2x = 4 − 2
−2x = 2
x = −1
$(x - 4)^2 = x^2 - 16$
$x^2 - 8x + 16 - x^2 = -16$
−8x = −16 − 16
−8x = −32
x = 4
$(x + 1)^2 = x^2 + 1$
$x^2 + 2x + 1 - x^2 = 1$
2x = 1 − 1
2x = 0
x = 0
Пожаулйста, оцените решение
