Question

Решить уравнение (x-4)^3-(x-7)^2(x+2)-27=0

Answer 1

Відповідь:

Вот ответ

Пояснення:

Answer 2

To solve the equation $(x-4)^3 - (x-7)^2(x+2) - 27 = 0$, we'll follow these steps:

Step 1: Expand the terms. Step 2: Combine like terms. Step 3: Solve for $x$.

Let's start with step 1:

Step 1: Expand the terms: $(x-4)^3 = (x-4)(x-4)(x-4) = (x^2 - 8x + 16)(x - 4) = x^3 - 12x^2 + 48x - 64$ $(x-7)^2 = (x-7)(x-7) = (x^2 - 14x + 49)$

Now, the equation becomes: $x^3 - 12x^2 + 48x - 64 - (x^2 - 14x + 49)(x + 2) - 27 = 0$

Step 2: Combine like terms:

$x^3 - 12x^2 + 48x - 64 - (x^3 + 2x^2 - 14x^2 - 28x + 49x + 98) - 27 = 0$

Simplify the equation further:

$x^3 - 12x^2 + 48x - 64 - (x^3 - 12x^2 + 21x - 98) - 27 = 0$

Now, the equation becomes:

$x^3 - 12x^2 + 48x - 64 - x^3 + 12x^2 - 21x + 98 - 27 = 0$

Step 3: Solve for $x$:

The common terms cancel out, leaving us with:

$-21x + 7 = 0$

Now, isolate $x$:

$21x = 7$

$x = \frac{7}{21}$

$x = \frac{1}{3}$

So, the solution to the equation is $x = \frac{1}{3}$.