Question

Уравнения:1)(х²+х+1)(х²+х+2)=122)3(х²+5х+1)²+2х²+10х=33)(х⁴-5х²)²-2(х⁴-5х²)=24​

Answer 1

1) (x² + x + 1)(x² + x + 2) = 12

Сделаем замену :

x² + x + 1 = m , тогда  x² + x + 2 = m + 1

m(m + 1) = 12

m² + m - 12 = 0

D = 1² - 4 * (- 12) = 1 + 48 = 49 = 7²

$m_{1}= \frac{-1-7}{2}=-4\\\\m_{2}=\frac{-1+7}{2}=3$

1) x² + x + 1 = - 4

x² + x + 5 = 0

D = 1² - 4 * 5 = 1 - 20 = - 19 < 0 - решений нет

2) x² + x + 1 = 3

x² + x - 2 = 0

По теореме Виета :

x₁ = - 2        x₂ = 1

Ответ : - 2 ; 1

2)

3(x² + 5x + 1)² + 2x² + 10x = 3

3(x² + 5x + 1)² + 2(x² + 5x) = 3

Сделаем замену :

x² + 5x + 1 = m , тогда x² + 5x = m - 1

3m² + 2(m - 1) = 3

3m² + 2m - 2 - 3 = 0

3m² + 2m - 5 = 0

D = 2² - 4 * 3 * (- 5) = 4 + 60 = 64 = 8²

$m_{1}=\frac{-2-8}{6}=-\frac{5}{3}\\\\m_{2} =\frac{-2+8}{6}=1$

1)x²+ 5x + 1 = - 5/3

x² + 5x + 8/3 = 0

3x² + 15x + 8 = 0

D = 15² - 4 * 3 * 8 = 225 - 96 = 129

$x_{1}=\frac{-15-\sqrt{129} }{6}\\\\x_{2}=\frac{-15+\sqrt{129} }{6}$

2)x²+ 5x + 1 = 1

x² + 5x = 0

x(x + 5) = 0

x₃ = 0         x₄ = - 5

3)

(x⁴ - 5x²)² - 2(x⁴ - 5x²) = 24

Сделаем замену :

x⁴ - 5x² = m

m² - 2m - 24 = 0

По теореме Виета :

m₁ = 6        m₂ = - 4

1) x⁴ - 5x² = 6

x⁴ - 5x² - 6 = 0

x² = 6

x₁ = - √6       x₂ = √6

x² = - 1  -   решений нет

2) x⁴ - 5x² = - 4

x⁴ - 5x² + 4 = 0

x² = 4

x₃ = - 2       x₄ = 2

x² = 1

x₅ = - 1      x₆ = 1

Ответ : - √6 ; √6 ; - 1 ; 1 ; - 2 ; 2

Answer 2

To solve these equations, we'll start by simplifying and then solving each one step by step.

1. (х²+х+1)(х²+х+2) = 12

Let's multiply the two quadratic expressions: (х²+х+1)(х²+х+2) = (х²+х)(х²+х+2) + (1)(х²+х+2) = х²(х²+х+2) + х(х²+х+2) + х²+х+2 = х⁴ + х³ + 2х² + х³ + х² + 2х + х² + х + 2 = х⁴ + 2х³ + 4х² + 3х + 2

Now, we have the equation: х⁴ + 2х³ + 4х² + 3х + 2 = 12

Rearrange the equation and set it equal to zero: х⁴ + 2х³ + 4х² + 3х + 2 - 12 = 0

Simplify further: х⁴ + 2х³ + 4х² + 3х - 10 = 0

Unfortunately, this is a fourth-degree polynomial equation, which doesn't have a simple analytical solution. We'll need to use numerical methods or software to find approximate solutions.

2. 3(х²+5х+1)² + 2х² + 10х = 3

First, expand the squared term: 3(х²+5х+1)² + 2х² + 10х = 3 3(х²+5х+1)(х²+5х+1) + 2х² + 10х = 3 3(х⁴ + 10х³ + 26х² + 10х + 1) + 2х² + 10х = 3 3х⁴ + 30х³ + 78х² + 30х + 3 + 2х² + 10х = 3 3х⁴ + 30х³ + 80х² + 40х = 0

Now, let's rearrange the equation and set it equal to zero: 3х⁴ + 30х³ + 80х² + 40х = 0

Divide the whole equation by 10 to simplify: х⁴ + 10х³ + 8х² + 4х = 0

Unfortunately, this equation is also not easily solvable analytically due to its fourth-degree nature. We need to use numerical methods or software to find approximate solutions.

3. (х⁴-5х²)² - 2(х⁴-5х²) = 24

Let's simplify the equation step by step:

(х⁴-5х²)² - 2(х⁴-5х²) = 24 (х⁴-5х²)(х⁴-5х²) - 2(х⁴-5х²) = 24 х⁸ - 10х⁶ + 25х⁴ - 2х⁴ + 10х² = 24 х⁸ - 10х⁶ + 23х⁴ + 10х² = 24

Now, set the equation equal to zero: х⁸ - 10х⁶ + 23х⁴ + 10х² - 24 = 0

Unfortunately, this is an eighth-degree polynomial equation, which is even more complex to solve analytically. We'll need to use numerical methods or software to find approximate solutions.

In all three cases, numerical methods (like Newton's method or software such as Mathematica, MATLAB, or Python's numerical libraries) can be used to find the approximate solutions to these equations.