Question

(4x-1)(x-2)+(2-x)(1-x)=0​

Answer 1

Ответ:

Раскрываем скобки и решаем два уравнения при равно 0.

х1 = 0,4

х2 = 2

Answer 2

Ответ:

Как то так.

Объяснение:

Answer 3

To solve the equation (4x-1)(x-2) + (2-x)(1-x) = 0, you can follow these steps:

1. Expand both terms on the left side of the equation:

   (4x-1)(x-2) + (2-x)(1-x) = 0

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

2. Combine like terms:

   4x^2 - 8x - x + 2 + 2 - x - x^2 + x = 0

3. Simplify the equation:

   4x^2 - 9x + 4 = 0

4. Now, you have a quadratic equation in the form of ax^2 + bx + c = 0, where a = 4, b = -9, and c = 4. To solve this quadratic equation, you can use the quadratic formula:

   x = (-b ± √(b² - 4ac)) / (2a)

   Plugging in the values:

   x = (-(-9) ± √((-9)² - 4(4)(4))) / (2(4))

   x = (9 ± √(81 - 64)) / 8

   x = (9 ± √17) / 8

So, the solutions for the equation (4x-1)(x-2) + (2-x)(1-x) = 0 are:

x = (9 + √17) / 8 x = (9 - √17) / 8