Question

(x-3)^x^2-x=(x-3)^7x-5

Answer 1

Ответ:

1-x4

___

x3

Пошаговое объяснение:

правильный ответ

Answer 2

To solve the equation $(x-3)^{x^2-x} = (x-3)^{7x-5}$, you can use the property of exponents that states: If $a^m = a^n$, where $a$ is a nonzero number, and $m$ and $n$ are real numbers, then $m = n$.

So, in your equation, you can set the exponents equal to each other:

$$x^2 - x = 7x - 5$$

Now, let's solve this equation for $x$:

$$x^2 - x - 7x + 5 = 0$$

Combine like terms:

$$x^2 - 8x + 5 = 0$$

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

$$x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$$

In this case, $a = 1$, $b = -8$, and $c = 5$. Plug these values into the quadratic formula:

$$x = \frac{-(-8) \pm \sqrt{(-8)^2 - 4(1)(5)}}{2(1)}$$

Simplify:

$$x = \frac{8 \pm \sqrt{64 - 20}}{2}$$

$$x = \frac{8 \pm \sqrt{44}}{2}$$

Now, you can simplify further:

$$x = \frac{8 \pm 2\sqrt{11}}{2}$$

Factor out a 2 from the numerator:

$$x = \frac{2(4 \pm \sqrt{11})}{2}$$

Now, cancel out the common factor of 2:

$$x = 4 \pm \sqrt{11}$$

So, the solutions to the equation $(x-3)^{x^2-x} = (x-3)^{7x-5}$ are $x = 4 + \sqrt{11}$ and $x = 4 - \sqrt{11}$.