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Задача по математике, 8 класс: Equations with a parameter. Пошаговые подсказки, полное решение и ответ. Длительность — 55 мин.
Условие задачи
Подсказки для объяснения
Открывайте подсказки постепенно — сначала дайте ребенку подумать самостоятельно.
Подсказка 1
Find the domain first: the denominator cannot be zero, so x ≠ -1.
Подсказка 2
Multiply both sides by x + 1, but remember that x = -1 is not allowed.
Подсказка 3
After simplification, express x in terms of a and check whether the obtained value satisfies the domain restriction.
Подсказка 4
Investigate special values of a that may make the equation impossible or turn it into an identity.
Пошаговое решение
Given (x - 2)/(x + 1) = a, with x ≠ -1. Multiply both sides by x + 1: x - 2 = a(x + 1) x - 2 = ax + a x - ax = a + 2 x(1 - a) = a + 2 If a ≠ 1, then x = (a + 2)/(1 - a) = -(a + 2)/(a - 1). Now check the restriction x ≠ -1. Substitute x = -1 into the equation: (-1 - 2)/(-1 + 1) = a This is undefined, so x = -1 is never a valid solution. We check whether the formula gives x = -1 for some a: -(a + 2)/(a - 1) = -1 a + 2 = a - 1, which is impossible. So the obtained solution is always valid when a ≠ 1. If a = 1, then the equation becomes (x - 2)/(x + 1) = 1, so x - 2 = x + 1, -2 = 1, which is impossible. Therefore, there are no solutions when a = 1. Conclusion: - one solution for every a ≠ 1; - no solutions for a = 1; - infinitely many solutions do not occur.
