Решите уравнение:
а) x(x + 1)(x − 10) = (x − 1)(x − 3)(x − 5);
б) $(x - 1)(x - 4)(x + 7) = x(x + 1)^2$.
x(x + 1)(x − 10) = (x − 1)(x − 3)(x − 5)
$(x^2 + x)(x - 10) = (x^2 - x - 3x + 3)(x - 5)$
$x^3 - 10x^2 + x^2 - 10x = (x^2 - 4x + 3)(x - 5)$
$x^3 - 9x^2 - 10x = x^3 - 4x^2 + 3x - 5x^2 + 20x - 15$
$x^3 - 9x^2 - 10x - x^3 + 9x^2 - 23x = -15$
−33x = −15
$x = \frac{15}{33} = \frac{5}{11}$
$(x - 1)(x - 4)(x + 7) = x(x + 1)^2$
$(x^2 - x - 4x + 4)(x + 7) = x(x + 1)(x + 1)$
$(x^2 - 5x + 4)(x + 7) = (x^2 + x)(x + 1)$
$x^3 - 5x^2 + 4x + 7x^2 - 35x + 28 = x^3 + x^2 + x^2 + x$
$x^3 + 2x^2 - 31x - x^3 - 2x^2 - x = -28$
−32x = −28
$x = \frac{28}{32} = \frac{7}{8}$
Пожауйста, оцените решение