To add the expression \(\frac{2x+1}{x-2} + \frac{x-3}{x^2} + (2x-8)\), we first need to find a common denominator and simplify where possible.
Step 1: Rewrite the expression
We have: \[ \frac{2x+1}{x-2} + \frac{x-3}{x^2} + (2x-8) \]
Step 2: Find a common denominator
The denominators are \(x-2\) and \(x^2\). The least common denominator (LCD) would be \(x^2(x-2)\).
Step 3: Rewrite each term with the common denominator
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For \(\frac{2x+1}{x-2}\): \[ \frac{2x+1}{x-2} = \frac{(2x+1)x^2}{x^2(x-2)} \]
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For \(\frac{x-3}{x^2}\): \[ \frac{x-3}{x^2} = \frac{(x-3)(x-2)}{x^2(x-2)} \]
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For \(2x-8\): \[ 2x - 8 = \frac{(2x - 8) x^2(x-2)}{x^2(x-2)} = \frac{(2x-8)x^2(x-2)}{x^2(x-2)} \]
Step 4: Combine all terms over the common denominator
Now we can add these fractions: \[ \frac{(2x+1)x^2 + (x-3)(x-2) + (2x-8)x^2(x-2)}{x^2(x-2)} \]
Step 5: Simplify the numerator
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Simplifying \((x-3)(x-2)\): \[ (x-3)(x-2) = x^2 - 5x + 6 \]
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Expanding \((2x-8)x^2(x-2)\): \[ (2x-8)x^2(x-2) = (2x-8)(x^3 - 2x^2) = 2x^4 - 4x^3 - 8x^2 + 16x \]
Now we combine everything in the numerator:
- From \((2x+1)x^2\): \[ 2x^3 + x^2 \]
- From \((x-3)(x-2)\): \[ x^2 - 5x + 6 \]
- From \((2x-8)x^2(x-2)\): \[ 2x^4 - 4x^3 - 8x^2 + 16x \]
Now the complete numerator gives us: \[ 2x^4 + (2x^3 - 4x^3) + (x^2 + x^2 - 8x^2) + (-5x + 16x + 6) \] Which simplifies to: \[ 2x^4 - 2x^3 - 6x^2 + 11x + 6 \]
Step 6: Final expression
Putting this all together, we have: \[ \frac{2x^4 - 2x^3 - 6x^2 + 11x + 6}{x^2(x-2)} \]
Conclusion
The simplest form of the sum \(\frac{2x+1}{x-2} + \frac{x-3}{x^2} + (2x-8)\) is: \[ \frac{2x^4 - 2x^3 - 6x^2 + 11x + 6}{x^2(x-2)} \]