Решите уравнение, используя введение новой переменной: а) (2х2 + З)2 – 12 ⋅ (2х2 + 3) + 11 = 0; б) (t2 – 2t)2 – 3 = 2(t2 – 2t); в) (x2 + x – 1) ⋅ (x2 + x + 2) = 40; г) (2x2 + x – 1) ⋅ (2x2 + x – 4) + 2 = 0.
Решение:
а) (2х2 + З)2 – 12 ⋅ (2х2 + 3) + 11 = 0
Замена 2x2 + 3 = t
t2 – 12t + 11 = 0
D = (-12)2 – 4 ⋅ 1 ⋅ 11 = 144 – 44 = 100 больше 0
√D = √100 = √102 = 10
t1 = -(-12) – 10 / 2 = 12 – 10 / 2 = 2/2 = 1
t2 = -(-12) + 10 / 2 = 12 + 10 / 2 = 22/2 = 11
t = 1
2x2 + 3 = 1
2x2 = 1 – 3
2x2 = -2
x2 = -1 нет корня
при t = 11
2x2 + 3 = 11
2x2 = 11 – 3
2x2 = 8
x2 =4
x = +-√4
x = +-2
б) (t2 – 2t)2 – 3 = 2(t2 – 2t)
Замена t2 – 2t = x
x2 – 3 – 2x = 0
x2 – 2x – 3 = 0
D = (-2)2 – 4 ⋅ 1 ⋅ (-3) = 4 + 12 = 16 больше 0
√D = √16 = √42 = 4
x1 = -(-2) – 4 / 2 = 2 – 4 / 2 = -2/2 = -1
x2 = -(-2) + 4 / 2 = 2 + 4 / 2 = 6/2 = 3
x = -1
t2 – 2t = -1
t2 – 2x + 1 = 0
(t – 1)2 = 0
t – 1 = 0
t = 1
При x = 3
t2 – 2t = 3
t2 – 2x – 3 = 0
D = (-2)2 – 4 ⋅ 1 ⋅ (-3) = 4 + 12 = 16 больше 0
√D = √16 = √42 = 4
t1 = -(-2) – 4 / 2 = 2 – 4 / 2 = -2/2 = -1
t1 = -(-2) + 4 / 2 = 2 + 4 / 2 = 6/2 = 3
в) (x2 + x – 1) ⋅ (x2 + x + 2) = 40
Замена x2 + x = t
(t – 1) ( t + 2) = 40
t2 + 2t – t – 2 – 40 = 0
t2 + t – 42 = 0
D = 12 – 4 ⋅ 1 ⋅ (-42) = 1 + 168 = 169 больше 0
√D = √169 = √132 = 13
t1 = – 1 – 13 / 2 = – ( 1 + 13) / 2 = – 14 / 2 = -7
t2 = – 1 + 13 / 2 = – ( 13 – 1) / 2 = 12 / 2 = 6
При t = – 7
x2 + x = -7
x2 + x + 7 = 0
D = 12 – 4 ⋅ 1 ⋅ 7 = 1 – 28 = – 27 больше 0 корней нет
√D = √25 = √52 = 5
x1 = – 1 – 5 / 2 = – (1 + 5) / 2 = – 6 / 2 = -3
x2 = – 1 + 5 / 2 = – (5 – 1) / 2 = 4 / 2 = 2
x = -3; x = 2
г) (2x2 + x – 1) ⋅ (2x2 + x – 4) + 2 = 0
Замена 2x2 + x = t
(t – 1) (t – 4) + 2 = 0
t2 – 4t – t + 4 + 2 = 0
t2 – 5t + 6 = 0
D = (-5)2 – 4 ⋅ 1 ⋅ 6 = 25 – 24 = 1 больше 0
√D = √1 = 1
t1 = -(-5) – 1 / 2 = 5 – 1 / 2 = 4 / 2 = 2
t2 = -(-5) + 1 / 2 = 5 + 1 / 2 = 6 / 2 = 3
При t = 2
2x2 + x = 2
2x2 + x – 2 = 0
D = 12 – 4 ⋅ 2 ⋅ (-2) = 1 + 16 = 17 больше 0
√D = √17
x1 = – 1 – √17 / 2 ⋅ 2 = – 1 – √17 / 4
x2 = – 1 + √17 / 2 ⋅ 2 = – 1 + √17 / 4
При t = 3
2x2 + x = 3
2x2 + x – 3 = 0
D = 12 – 4 ⋅ 2 ⋅ (-3) = 1 + 24 = 25 больше 0
√D = √25 = √52 = 5
x1 = -1 – 5 / 2 ⋅ 2 = – ( 1 + 5) / 4 = – 6/4 = -1,5
x2 = -1 + 5 / 2 ⋅ 2 = 5 – 1 / 4 = 4/4 = 1
x = – 1 +- √17 / 4; x = -1,5; x = 1
Источник : Учебник по Алгебре 9 класса авторов Ю.Н. Макарычев, Н.Г. Миндюк, К.И. Нешков, С.Б. Суворова, 2014г.