Sure! Let's find the simplest form of the expression [tex]\(2m - n\)[/tex] given the definitions of [tex]\(m\)[/tex] and [tex]\(n\)[/tex].
First, let's rewrite [tex]\(m\)[/tex] and [tex]\(n\)[/tex] using their given expressions:
[tex]\[ m = \frac{2x}{1 - x^2} \][/tex]
[tex]\[ n = \frac{2x}{1 + x} \][/tex]
Now, we need to compute [tex]\(2m - n\)[/tex]:
[tex]\[ 2m - n = 2 \left( \frac{2x}{1 - x^2} \right) - \frac{2x}{1 + x} \][/tex]
Calculate [tex]\(2m\)[/tex]:
[tex]\[ 2m = 2 \left( \frac{2x}{1 - x^2} \right) = \frac{4x}{1 - x^2} \][/tex]
Now substitute [tex]\(2m\)[/tex] and [tex]\(n\)[/tex] back into the expression:
[tex]\[ 2m - n = \frac{4x}{1 - x^2} - \frac{2x}{1 + x} \][/tex]
To combine these fractions, we need a common denominator. The common denominator for [tex]\(\frac{4x}{1 - x^2}\)[/tex] and [tex]\(\frac{2x}{1 + x}\)[/tex] is [tex]\((1 - x^2)(1 + x)\)[/tex].
First, notice that:
[tex]\[ 1 - x^2 = (1 - x)(1 + x) \][/tex]
Thus we can rewrite:
[tex]\[ \frac{4x}{1 - x^2} = \frac{4x}{(1 - x)(1 + x)} \][/tex]
To convert this to a common denominator, we'll use [tex]\((1 - x)(1 + x)\)[/tex] for both terms:
[tex]\[ \frac{4x}{(1 - x)(1 + x)} - \frac{2x}{1 + x} \cdot \frac{1 - x}{1 - x} \][/tex]
This simplifies to:
[tex]\[ \frac{4x}{(1 - x)(1 + x)} - \frac{2x(1 - x)}{(1 - x)(1 + x)} \][/tex]
Combine the two fractions:
[tex]\[ \frac{4x - 2x(1 - x)}{(1 - x)(1 + x)} \][/tex]
Expand the numerator:
[tex]\[ 4x - 2x + 2x^2 = 2x + 2x^2 \][/tex]
Combine like terms:
[tex]\[ \frac{2x + 2x^2}{(1 - x)(1 + x)} \][/tex]
We can factor out a 2x from the numerator:
[tex]\[ \frac{2x(1 + x)}{(1 - x)(1 + x)} \][/tex]
Now notice that [tex]\((1 + x)\)[/tex] cancels out in the numerator and denominator:
[tex]\[ \frac{2x \cancel{(1 + x)}}{(1 - x)\cancel{(1 + x)}} = \frac{2x}{1 - x} \][/tex]
So, the simplest form of [tex]\(2m - n\)[/tex] is:
[tex]\[ 2m - n = \frac{2x}{1 - x} \][/tex]
However, due to a negative sign simplification, the correct answer in simplest form is:
[tex]\[ 2m - n = -\frac{2x}{x - 1} \][/tex]
