Ответ учителя по предмету Алгебра
1)
x^3 + 2x^2+x = 0;
x(x^2 + 2x + 1) = 0;
Т.к. произведение равно нулю, то хотя бы один из множителей равен нулю:
I: x = 0;
II: x^2 + 2x + 1 = 0;
x1 = (-2+√4-4)/2
x1 = -1;
x2 = (-2-√4-4)/2
x2 = -1
Ответ: х = 0 ; x = -1;
2)
x/(x-2) = 12(x+2);
Применяем метод пропорции:
x^2 + 2x = 12x — 24
x^2 + 2x — 12x = -24
x^2 — 10x + 24 = 0;
x1 = (10+√100-4*24)/2
x1 = 14/2
x1 = 7
x2 = (10-√100-4*24)/2
x2 = 6/2
x2 = 3
Ответ: x = 7
x = 3
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