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maibui2011
@maibui2011
July 2022
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20 БАЛЛОВ! ПОМОГИТЕ РЕШИТЬ ПЕРВОЕ УРАВНЕНИЕ ПОЖАЛУЙСТА!!!!!!!!
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Удачник66
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1. Это так называемое симметричное (или симметрическое) уравнение.
(x^2 + 1/x^2) + 6*(x - 1/x) + 6 = 0
(x - 1/x)^2 = x^2 - 2*x*1/x + 1/x^2 = x^2 + 1/x^2 - 2
(x^2 - 2 + 1/x^2) + 6*(x - 1/x) + 8 = 0
Замена (x - 1/x) = y
y^2 + 6y + 8 = 0
(y + 2)(y + 4) = 0
1) y = x - 1/x = -2
x^2 + 2x - 1 = 0
D = 4 + 4 = 8 = (2√2)^2
x1 = (-2 - 2√2)/2 = -1 - √2; x2 = -1 + √2
2) y = x - 1/x = -4
x^2 + 4x - 1 = 0
D = 16 + 4 = 20 = (2√5)^2
x3 = (-4 - 2√5)/2 = -2 - √5; x4 = -2 + √5
Всё!
2. Тут проще. Замена y = x^2 - 3x + 4
1/(y - 1) + 2/y = 6/(y + 1)
Общий знаменатель y(y - 1)(y + 1)
y(y + 1) + 2(y - 1)(y + 1) = 6y(y - 1)
y^2 + y + 2y^2 - 2 = 6y^2 - 6y
0 = 3y^2 - 7y + 2
D = 7^2 - 4*3*2 = 49 - 24 = 25 = 5^2
1) y = x^2 - 3x + 4 = (7 - 5)/6 = 2/6 = 1/3
3x^2 - 9x + 12 - 1 = 0
D = 9^2 - 4*11*3 = 81 - 132 < 0
Решений нет
2) y = x^2 - 3x + 4 = (7 + 5)/6 = 2
x^2 - 3x + 2 = 0
(x - 1)(x - 2) = 0
x1 = 1; x2 = 2
2 votes
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Answers & Comments
Verified answer
1. Это так называемое симметричное (или симметрическое) уравнение.(x^2 + 1/x^2) + 6*(x - 1/x) + 6 = 0
(x - 1/x)^2 = x^2 - 2*x*1/x + 1/x^2 = x^2 + 1/x^2 - 2
(x^2 - 2 + 1/x^2) + 6*(x - 1/x) + 8 = 0
Замена (x - 1/x) = y
y^2 + 6y + 8 = 0
(y + 2)(y + 4) = 0
1) y = x - 1/x = -2
x^2 + 2x - 1 = 0
D = 4 + 4 = 8 = (2√2)^2
x1 = (-2 - 2√2)/2 = -1 - √2; x2 = -1 + √2
2) y = x - 1/x = -4
x^2 + 4x - 1 = 0
D = 16 + 4 = 20 = (2√5)^2
x3 = (-4 - 2√5)/2 = -2 - √5; x4 = -2 + √5
Всё!
2. Тут проще. Замена y = x^2 - 3x + 4
1/(y - 1) + 2/y = 6/(y + 1)
Общий знаменатель y(y - 1)(y + 1)
y(y + 1) + 2(y - 1)(y + 1) = 6y(y - 1)
y^2 + y + 2y^2 - 2 = 6y^2 - 6y
0 = 3y^2 - 7y + 2
D = 7^2 - 4*3*2 = 49 - 24 = 25 = 5^2
1) y = x^2 - 3x + 4 = (7 - 5)/6 = 2/6 = 1/3
3x^2 - 9x + 12 - 1 = 0
D = 9^2 - 4*11*3 = 81 - 132 < 0
Решений нет
2) y = x^2 - 3x + 4 = (7 + 5)/6 = 2
x^2 - 3x + 2 = 0
(x - 1)(x - 2) = 0
x1 = 1; x2 = 2