Let's start with the expressions for [tex]\( m \)[/tex] and [tex]\( n \)[/tex]:
[tex]\[ m = \frac{2x}{1 - x^2} \][/tex]
[tex]\[ n = \frac{2x}{1 + x} \][/tex]
We aim to find [tex]\( 2m - n \)[/tex]. Substitute the given expressions for [tex]\( m \)[/tex] and [tex]\( n \)[/tex] into the expression:
[tex]\[ 2m - n = 2 \left( \frac{2x}{1 - x^2} \right) - \frac{2x}{1 + x} \][/tex]
Let's break it down, step by step:
1. First, calculate [tex]\( 2m \)[/tex]:
[tex]\[ 2m = 2 \cdot \frac{2x}{1 - x^2} = \frac{4x}{1 - x^2} \][/tex]
2. Now, subtract [tex]\( n \)[/tex] from [tex]\( 2m \)[/tex]:
[tex]\[ 2m - n = \frac{4x}{1 - x^2} - \frac{2x}{1 + x} \][/tex]
To combine these fractions, we need a common denominator. The denominators here are [tex]\( 1 - x^2 \)[/tex] and [tex]\( 1 + x \)[/tex]. Notice that:
[tex]\[ 1 - x^2 = (1 + x)(1 - x) \][/tex]
Thus, our common denominator will be [tex]\( (1 + x)(1 - x) \)[/tex]. Rewrite the fractions with this common denominator:
For the first term:
[tex]\[ \frac{4x}{1 - x^2} = \frac{4x}{(1 + x)(1 - x)} \][/tex]
For the second term:
[tex]\[ \frac{2x}{1 + x} = \frac{2x(1 - x)}{(1 + x)(1 - x)} = \frac{2x - 2x^2}{(1 + x)(1 - x)} \][/tex]
Now we can combine them under a single denominator:
[tex]\[ 2m - n = \frac{4x - (2x - 2x^2)}{(1 + x)(1 - x)} \][/tex]
Simplify the numerator:
[tex]\[ 4x - (2x - 2x^2) = 4x - 2x + 2x^2 = 2x (1 + x) \][/tex]
Therefore:
[tex]\[ 2m - n = \frac{2x(1 + x)}{(1 + x)(1 - x)} = \frac{2x}{1 - x} \][/tex]
Our simplified expression is:
[tex]\[ 2m - n = \frac{-2x}{x - 1} \][/tex]
This can be further simplified to:
[tex]\[ 2m - n = \frac{-2x}{x - 1} \][/tex]
So, the simplest form of [tex]\( 2m - n \)[/tex] in terms of [tex]\( x \)[/tex] is:
[tex]\[ \boxed{\frac{-2x}{x-1}} \][/tex]
