Let's solve the given mathematical problem step by step to simplify the expression:
[tex]\[
\frac{\frac{x-8}{x+8}+\frac{x-1}{x+1}}{\frac{x}{x+1}-\frac{5x-1}{x}}
\][/tex]
Step 1: Simplify the numerator
Numerator: [tex]\(\frac{x-8}{x+8} + \frac{x-1}{x+1}\)[/tex].
To add these fractions, we need a common denominator:
- The common denominator here is [tex]\((x + 8)(x + 1)\)[/tex].
Rewriting each fraction with the common denominator:
[tex]\[
\frac{x-8}{x+8} = \frac{(x-8)(x+1)}{(x+8)(x+1)}
\][/tex]
[tex]\[
\frac{x-1}{x+1} = \frac{(x-1)(x+8)}{(x+8)(x+1)}
\][/tex]
Now add the fractions:
[tex]\[
\frac{(x-8)(x+1) + (x-1)(x+8)}{(x+8)(x+1)}
\][/tex]
Expanding the numerators:
[tex]\[
(x-8)(x+1) = x^2 + x - 8x - 8 = x^2 - 7x - 8
\][/tex]
[tex]\[
(x-1)(x+8) = x^2 + 8x - x - 8 = x^2 + 7x - 8
\][/tex]
Adding these together:
[tex]\[
x^2 - 7x - 8 + x^2 + 7x - 8 = 2x^2 - 16
\][/tex]
So the numerator is:
[tex]\[
\frac{2x^2 - 16}{(x + 8)(x + 1)} = \frac{2(x^2 - 8)}{(x + 8)(x + 1)}
\][/tex]
Step 2: Simplify the denominator
Denominator: [tex]\(\frac{x}{x+1} - \frac{5x-1}{x}\)[/tex].
To subtract these fractions, we need a common denominator:
- The common denominator here is [tex]\(x(x+1)\)[/tex].
Rewriting each fraction with the common denominator:
[tex]\[
\frac{x}{x+1} = \frac{x \cdot x}{x(x+1)} = \frac{x^2}{x(x+1)}
\][/tex]
[tex]\[
\frac{5x-1}{x} = \frac{(5x-1)(x+1)}{x(x+1)}
\][/tex]
Now subtract the fractions:
[tex]\[
\frac{x^2 - (5x-1)(x+1)}{x(x+1)}
\][/tex]
Expanding the numerators:
[tex]\[
(5x-1)(x+1) = 5x^2 + 5x - x - 1 = 5x^2 + 4x - 1
\][/tex]
Subtracting:
[tex]\[
x^2 - (5x^2 + 4x - 1) = x^2 - 5x^2 - 4x + 1 = -4x^2 - 4x + 1
\][/tex]
So the denominator is:
[tex]\[
\frac{-4x^2 - 4x + 1}{x(x + 1)}
\][/tex]
Step 3: Combine the simplified numerator and denominator
[tex]\[
\frac{\frac{2(x^2 - 8)}{(x + 8)(x + 1)}}{\frac{-4x^2 - 4x + 1}{x(x + 1)}}
\][/tex]
To divide fractions, multiply by the reciprocal:
[tex]\[
\frac{2(x^2 - 8)}{(x + 8)(x + 1)} \cdot \frac{x(x + 1)}{-4x^2 - 4x + 1}
\][/tex]
Simplify this expression by canceling out common factors:
[tex]\[
\frac{2(x^2 - 8) \cdot x(x + 1)}{(x + 8)(x + 1)(-4x^2 - 4x + 1)}
\][/tex]
The [tex]\((x + 1)\)[/tex] terms cancel out:
So we are left with:
[tex]\[
\frac{2x(x^2 - 8)}{(x + 8)(-4x^2 - 4x + 1)}
\][/tex]
In a more factored form, this is:
[tex]\[
\frac{2x(x^2 - 8)}{(x + 8)(-4x^2 - 4x + 1)}
\][/tex]
Therefore, the expression in factored form is:
[tex]\[
\boxed{\frac{2x(x^2 - 8)}{4x^3 + 36x^2 + 31x - 8}}
\][/tex]
