Найдите корни уравнения: а) (3x – 2 / x – 1) – (2x – 3 / x + 3) = (12x + 4 / x2 + 2x – 3); б) (5x + 1 / x + 7) – (2x + 2 / x – 3) + (63 / x2 + 4x – 21) = 0; в) (x / x2 + 4x + 4) = (4 / x2 – 4) – (16 / x3 + 2x2 – 4x – 8).
Решение:
а) (3x – 2 / x – 1) – (2x – 3 / x + 3) = (12x + 4 / x2 + 2x – 3)
(3x – 2) (x + 3) – (2x + 3) ( x – 1) = 12x +4
3x2 + 9x – 2x – 6 – (2x2 – 2x + 3x – 3) = 12x + 4
3x2 + 7x – 6 – 2x2 – x + 3 – 12x – 4 = 0
x2 – 6x – 7 = 0
x1 ⋅ x2 = -7
x1 + x2 = 6
x1 = – 1
x2 = 7
Тогда x = -1
x2 + 2x – 3 = (-1)2 + 2 ⋅ (-1) – 3 = 1 – 2 – 3 = -4
Тогда x = 7
x2 + 2x – 3 = 72 + 2 ⋅ 7 – 3 = 49 + 14 – 3 = 60
x = 1; x = 7
б) (5x + 1 / x + 7) – (2x + 2 / x – 3) + (63 / x2 + 4x – 21) = 0
(5x – 1) (x – 3) – (2x + 2) (x + 7) + 63 = 0
5x2 – 15x – x + 3 – (2x2 + 2x + 14x + 14) + 63 = 0
5x2 – 16x + 3 – 2x2 – 16x – 14 + 63 = 0
3x2 – 32x + 52 = 0
D = (-32)2 – 4 ⋅ 3 ⋅ 52 = 1024 – 624 = 400 больше 0
√D = √400 = √202 = 20
x1 = – (-32) + 20 / 2 ⋅ 3 = 32 – 20 / 6 = 12 / 6 = 2
x2 = – (-32) + 20 / 2 ⋅ 3 = 32 + 20 / 6 = 52 / 6 = 8 2/3
Тогда x = 2
x2 + 4x – 21 = 22 + 4 ⋅ 2 – 21 = 4 + 8 – 21 = – 9
Тогда x = 8 2/3
x2 + 4x – 21 = (x + 7) (x – 3) = (8 2/3 + 7 ) ( 8 2/3 – 3) = 15 2/3 ⋅ 5 2/3 не равно 0
x = 2; x = 8 2/3
в) (x / x2 + 4x + 4) = (4 / x2 – 4) – (16 / x3 + 2x2 – 4x – 8)
x / x + 22 = 4 / x – 2) ( x +2) – 16 / x + 22 x – 2
x (x – 2) = 4 (x + 2) – 16
x2 – 2x = 4x + 8 – 16
x2 – 2x – 4x – 8 + 16 = 0
x2 – 6x + 8 = 0
x1 ⋅ x2 = 8
x1 + x2 = 6
x1 = 2
x2 = 4
Тогда x = 2
(x + 2)2 (x – 2) = (2 + 2)2 (2 – 2) = 42 ⋅ 0 = 16 ⋅ 0 = 0
Тогда x = 4
(x + 2)2 (x – 2) = (4 + 2)2 (4 – 2) = 62 ⋅ 2 = 36 ⋅ 2 = 72
x = 4
Ответ: а) x = 1, x = 7 ; б) x = 2, x = 8 2/3 ; в) x = 4
Источник : Учебник по Алгебре 9 класса авторов Ю.Н. Макарычев, Н.Г. Миндюк, К.И. Нешков, С.Б. Суворова, 2014г.