Question

(x-5)^6+(x^2-6x+7)^3=0

Answer 1

(x-5)⁶ + (x²-6x+7)³ = 0

Для начала решим дискриминант:

x²-6x+7 = 0

a=1 b=-6 c=7

D = b²-4ac => (-6)²-4*1*7 = 36-28 = 8 > 0, 2 корня.

x = $\frac{-b+-\sqrt{D} }{2a}$

x₁ = $\frac{6+8}{2} = 7$

x₂ = $\frac{6-8}{2} = -1$

Получаем:

(x-5)⁶ + ((x-7)(x+1))³

Или же:

(x-5) (x-7)(x+1)

x=5 или x=7 или x=-1

Мы решили дискриминант, чтобы найти корни уравнения. Теперь:

5⁶ + 7³ * (-1)³ = 5⁶ + (-7)⁶ = -2⁶ = -64 (вспомним правила степеней).

Возможно не совсем так. Если что-то не так, поправьте.

Answer 2

To solve the equation $(x-5)^6 + (x^2-6x+7)^3 = 0$, we can use a substitution to simplify the equation. Let's substitute $y = (x-5)^6$ and $z = (x^2-6x+7)^3$.

Using these substitutions, we have $y + z = 0$. Now, we can solve for $y$ and $z$ separately and then substitute back to find the values of $x$ that satisfy the original equation.

Solving for $y$: $y = (x-5)^6$

Taking the sixth root of both sides: $\sqrt[6]{y} = x - 5$

Adding 5 to both sides: $x = \sqrt[6]{y} + 5$

Solving for $z$: $z = (x^2-6x+7)^3$

Expanding the expression: $z = (x^2-6x+7)(x^2-6x+7)(x^2-6x+7)$

Let's simplify this equation by substituting $a = x^2-6x+7$: $z = a^3$

Now, we have $x = \sqrt[6]{y} + 5$ and $z = a^3$, where $y$ and $a$ are related to each other through the original equation.

Substituting back into the equation $y + z = 0$, we have: $\sqrt[6]{y} + 5 + a^3 = 0$

At this point, we need additional information to solve the equation. The original equation is missing a relationship between $y$ and $a$, so we cannot find a specific solution without more information.

If you have any additional constraints or relationships between $y$ and $a$, please provide them, and I'll be happy to assist you further.