Rewrite the expression in standard form.

[tex]\[
tima \, vi \, aue \, valar \, se \, le \, dibe \, Dar \, A \, \times \, \text{para} \, \text{Producir} \, \text{in}
\][/tex]

A) [tex]\(\frac{6x}{x-2}\)[/tex]

B) [tex]\(\frac{x^2+3x}{x^2+2x-3}\)[/tex]



Answer :

To determine the correct option in the given question, we need to compare the two expressions to see which one is equal to the expression presented. Here are the two expressions provided:

Option A:
[tex]\[ \frac{6x}{x-2} \][/tex]

Option B:
[tex]\[ \frac{x^2 + 3x}{x^2 + 2x - 3} \][/tex]

First, let's analyze Option B and simplify it to see if it matches Option A.

1. Factor the numerator and the denominator in Option B:

[tex]\[ \frac{x^2 + 3x}{x^2 + 2x - 3} \][/tex]

Numerator:
The numerator [tex]\(x^2 + 3x\)[/tex] can be factored as:
[tex]\[ x^2 + 3x = x(x + 3) \][/tex]

Denominator:
The denominator [tex]\(x^2 + 2x - 3\)[/tex] can be factored by finding the roots:
[tex]\[ x^2 + 2x - 3 = (x - 1)(x + 3) \][/tex]

2. Write the expression in its factored form:
[tex]\[ \frac{x(x + 3)}{(x - 1)(x + 3)} \][/tex]

3. Simplify the fraction:
We can cancel out the common factor [tex]\((x + 3)\)[/tex] in the numerator and the denominator:
[tex]\[ \frac{x(x + 3)}{(x - 1)(x + 3)} = \frac{x}{x - 1} \][/tex]

4. Compare the simplified form of Option B to Option A:
- Option A: [tex]\( \frac{6x}{x - 2} \)[/tex]
- Simplified Option B: [tex]\( \frac{x}{x - 1} \)[/tex]

Clearly, after simplification, Option B does not match with Option A.

Since we don't have to do any further analysis based on the above comparison, we conclude that Option A: [tex]\(\frac{6x}{x - 2}\)[/tex] and Option B: [tex]\(\frac{x}{x - 1}\)[/tex] are different from each other.

Thus, the correct option is:

[tex]\[ \boxed{2} \][/tex]