Question

4(2-х)+х(х-2)²+(2-х)³=0​

Answer 1

$8 - 4x + x{}^{3} - 4x {}^{2} + 4x + 8 - 12x + 6x {}^{2} - x {}^{3} = 0$

$16 + 2x {}^{2} - 12x = 0$

$8 + x {}^{2} - 6x = 0$

$x {}^{2} - 6x + 8 = 0$

$x {}^{2} - 2x - 4x + 8 = 0$

$x \times (x - 2) - 4(x - 2) = 0$

$(x - 2) \times (x - 4) = 0$

$x1 = 2 \\ x2 = 4$

Answer 2

Ответ: х1 = 2

х2 = 4

Объяснение:

Раскроем скобки:

8 - 4х + х(х² - 4х + 4) + 8 - 3•4•х + 3•2•х² - х³ = 0

8 -4х + х³ - 4х² +4х + 8 -12х + 6х² - х³=0

2х² - 12х + 16 = 0

Решим квадратное уравнение:

D = b² - 4ac = (-12)2 - 4·2·16 = 144 - 128 = 16

х1 = (12-4)÷4 = 2

х2 = (12+4)÷4 = 4

Answer 3

To solve the equation 4(2 - x) + x(x - 2)² + (2 - x)³ = 0, you can follow these steps:

1. Expand and simplify the equation:

   4(2 - x) + x(x - 2)² + (2 - x)³ = 0

   8 - 4x + x(x² - 4x + 4) + (2 - x)(2 - x)(2 - x) = 0

2. Continue simplifying:

   8 - 4x + x³ - 4x² + 4x + (2 - x)(2 - x)(2 - x) = 0

3. Expand (2 - x)(2 - x)(2 - x):

   (2 - x)(2 - x)(2 - x) = (2 - x)³

   (2 - x)³ = 8 - 12x + 6x² - x³

4. Replace (2 - x)³ in the equation:

   8 - 4x + x³ - 4x² + 4x + (8 - 12x + 6x² - x³) = 0

5. Group like terms and simplify:

   8 + 8 - 4x + 4x - 12x + 6x² + x³ - x³ = 0

6. The x³ terms cancel out:

   16 - 12x + 6x² = 0

7. Divide the entire equation by 2 to simplify:

   8 - 6x + 3x² = 0

8. Now, you have a quadratic equation. Let's solve it for x:

   3x² - 6x + 8 = 0

   To solve this quadratic equation, you can use the quadratic formula:

   x = (-b ± √(b² - 4ac)) / (2a)

   In this case, a = 3, b = -6, and c = 8.

   x = (-(-6) ± √((-6)² - 4(3)(8))) / (2(3))

   x = (6 ± √(36 - 96)) / 6

   x = (6 ± √(-60)) / 6

Since the discriminant (the value inside the square root) is negative, there are no real solutions to this equation. The solutions are complex numbers.

So, the solutions are:

x = (6 + √(-60)) / 6 and x = (6 - √(-60)) / 6

You can further simplify the expressions, but they will involve imaginary numbers.