At Westonci.ca, we provide clear, reliable answers to all your questions. Join our vibrant community and get the solutions you need. Get detailed answers to your questions from a community of experts dedicated to providing accurate information. Get detailed and accurate answers to your questions from a dedicated community of experts on our Q&A platform.
Sagot :
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]
[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]
Thanks for using our service. We aim to provide the most accurate answers for all your queries. Visit us again for more insights. Thank you for your visit. We're dedicated to helping you find the information you need, whenever you need it. We're glad you chose Westonci.ca. Revisit us for updated answers from our knowledgeable team.
