помогите решить[tex](\frac{x^2+6}{x})^2+2(\frac{x^2+6}{x})=35[/tex]. Created by irynasokolova. algebra-ru

Ответ:обозначимответ3. 2. -1. -6

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oqilovmehrob

Ответ:

обозначим

$\frac{ {x}^{2} + 6 }{x} = y \\ {y}^{2} + 2y - 35 = 0 \\ d = 4 + 140 = 144 \\ y1 = \frac{ - 2 + 12}{2} = 5 \\ y2 = \frac{ - 2 - 12}{2} = - 7$

$\frac{ {x}^{2} + 6 }{x} = 5 \\ {x}^{2} + 6 = 5x \\ {x}^{2} - 5x + 6 = 0 \\ d = 25 - 24 = 1 \\ x1 = \frac{5 + 1}{2} = 3 \\ x2 = \frac{5 - 1}{2} = 2$

$\frac{ {x}^{2} + 6 }{x} = - 7 \\ {x}^{2} + 6 = - 7x \\ {x}^{2} + 7x + 6 = 0 \\ d = 49 - 24 = 25 \\ x3 = \frac{ - 7 + 5}{2} = - 1 \\ x4 = \frac{ - 7 - 5}{2} = - 6$

ответ

3. 2. -1. -6

3 votes Thanks 2

olga0olga76

Ответ:

x1 = 3, x2 = 2, x3=-1, x4=-6

Объяснение:

пусть (х²+6)/х = а,

тогда а² + 2а = 35

а² + 2а - 35 = 0

D = (2)² -4*1*(-35) = 4+140 = 144 = 12² > 0

a1 = (-2 + 12) / (2*1) = 10 / 2 = 5

a2 = (-2 - 12) / (2*1) = -14 / 2 = -7

Выход из замены:

1) а1 = 5 = (х²+6)/х (ОДЗ : х ≠ 0)

(х²+6)/х = 5 или х²+6 = 5х (х ≠ 0)

х² - 5х +6 = 0

D = (-5)² - 4*1*6 = 25-24 = 1 = 1² > 0

x1 = (-(-5) + 1) / (2*1) = 6 / 2 =3

x2 = (-(-5) - 1) / (2*1) = 4 / 2 =2

2) а1 = -7 = (х²+6)/х (ОДЗ : х ≠ 0)

(х²+6)/х = -7 или х²+6 = -7х (х ≠ 0)

х² + 7х +6 = 0

D = (7)² - 4*1*6 = 49-24 = 25 = 5² > 0

x3= (-(7) + 5) / (2*1) = -2 / 2 = -1

x4 = (-(7) - 5) / (2*1) = -12 / 2 = -6

0 votes Thanks 0

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