Question

(2х-1)³+(2х²-5х+2)³=0

Answer 1

Объяснение:

$(2x-1)^3+(2x^2-5x+2)^3=0\\(2x-1)^3=-(2x^2-5x+2)^3\\(2x-1)^3=(-2x^2+5x-2)^3\\2x-1=-2x^2+5x-2\\2x^2-3x+1=0\\D=1\ \ \ \ \sqrt{D}=1\\ x_1=0,5\ \ \ \ x_2=1.$

Ответ: x₁=0,5, x₂=1.

Answer 2

Ответ:

(см. объяснение)

Объяснение:

$(2x-1)^3+(2x^2-5x+2)^3=0$

Воспользуемся формулой:

$a^3+b^3=(a+b)(a^2-ab+b^2)$

Значит исходное уравнение примет вид:

$(2x-1+2x^2-5x+2)((2x-1)^2-(2x-1)(2x^2-5x+2)+(2x^2-5x+2)^2)=0$

После упрощения получим:

$(2x^2-3x+1)(4x^4-24x^3+49x^2-33x+7)=0$

Попробуем разложить каждую скобку на множители.

С 1-ой все просто:

$2x^2-3x+1=2x^2-2x-x+1=2x(x-1)-(x-1)=(x-1)(2x-1)$

На ней подробно останавливаться не буду.

Рассмотрим теперь вторую скобку:

$4x^4-24x^3+49x^2-33x+7=\\=(4x^4-20x^3+28x^2)-(4x^3-20x^2+28x)+(x^2-5x+7)=\\=4x^2(x^2-5x+7)-4x(x^2-5x+7)+(x^2-5x+7)=(x^2-5x+7)(4x^2-4x+1)=\\=(2x-1)^2(x^2-5x+7)$

Так, мы разложили на множители вторую скобку.

Теперь запишем то, что у нас получилось:

$(x-1)(2x-1)(2x-1)^2(x^2-5x+7)=0\\(2x-1)^3(x-1)(x^2-5x+7)=0$

Теперь без труда найдем корни уравнения:

$\left[\begin{array}{c}x=1\\x=\dfrac{1}{2}\end{array}\right;$

Уравнение решено!

Answer 3

It seems like you've provided an equation with expressions involving the variable "x." You're looking for solutions to this equation. Let's solve it step by step.

The equation you've provided is: (2x - 1)³ + (2x² - 5x + 2)³ = 0

We can start by factoring the two cubes, and then we'll look for solutions.

Factor the first cube: (2x - 1)³ = 8x³ - 12x² + 6x - 1

Factor the second cube: (2x² - 5x + 2)³ = 8x⁶ - 60x⁵ + 150x⁴ - 120x³ + 36x² - 20x + 8

Now the equation becomes: 8x³ - 12x² + 6x - 1 + 8x⁶ - 60x⁵ + 150x⁴ - 120x³ + 36x² - 20x + 8 = 0

Combine like terms: 8x⁶ - 60x⁵ + 150x⁴ - 112x³ + 24x² - 14x + 7 = 0

Unfortunately, this equation doesn't seem to have a simple, closed-form solution. Solving higher-degree polynomial equations like this often involves numerical methods or approximation techniques. If you're looking for specific values of "x" that satisfy the equation, you might need to use a numerical solver or software to find approximate solutions.

If you have a specific goal in mind or any constraints on the solutions you're seeking, please provide more context so I can assist you further.