Question

X^-x-2/x-3=2x-4/x^2-3x​

Answer 1

х^2-х-2 = 0; Д=(-1)^2-4*1*(-2)=1+8=9; х1=(1+3)/2=2, х2=(1-3)/2=-1. Следовательно: х^2-х-2 = (х-2 )(х+1). Сокрашаем дроьь:

Answer 2

To solve the given equation $\frac{X^{-x}-2}{x-3} = \frac{2x-4}{x^2-3x}$, we need to find the value(s) of $x$ that satisfy the equation.

Let's start by simplifying the expressions on both sides of the equation.

Step 1: Simplify the left-hand side (LHS):

$\frac{X^{-x}-2}{x-3}$

To simplify this, we can rewrite $X^{-x}$ as $\frac{1}{X^x}$:

$\frac{\frac{1}{X^x}-2}{x-3}$

Step 2: Simplify the right-hand side (RHS):

$\frac{2x-4}{x^2-3x}$

Factor out $2$ from the numerator:

$\frac{2(x-2)}{x(x-3)}$

Now, our equation becomes:

$\frac{\frac{1}{X^x}-2}{x-3} = \frac{2(x-2)}{x(x-3)}$

Step 3: Get rid of the denominators by multiplying both sides by $(x-3)$ and $x(x-3)$:

$(x-3) \cdot \frac{\frac{1}{X^x}-2}{x-3} = x(x-3) \cdot \frac{2(x-2)}{x(x-3)}$

The $(x-3)$ term on the left side cancels out, and the $x(x-3)$ term on the right side cancels out:

$\frac{1}{X^x} - 2 = 2(x-2)$

Step 4: Eliminate the fraction by multiplying both sides by $X^x$:

$X^x \cdot \left(\frac{1}{X^x} - 2\right) = X^x \cdot (2(x-2))$

The first term on the left side cancels out, and the second term on the right side can be expanded:

$1 - 2X^x = 2x - 4$

Step 5: Move all the $X^x$ terms to one side and all the $x$ terms to the other side:

$-2X^x - 2x = -3$

Step 6: Divide through by $-1$ to simplify:

$2X^x + 2x = 3$

Step 7: Now, it might not be possible to solve this equation algebraically. If you have a specific value for $X$ (a constant) given in the original equation, you can plug it in and solve for $x$. However, if $X$ is also a variable, then it might not be possible to obtain an explicit solution for $x$ in terms of $X$.

Please check the original equation to see if there are any additional constraints or specific values for $X$ or $x$ that might help in finding a solution. Otherwise, this equation might not be solvable in general.