To simplify the expression \(\frac{3}{x^2 - 3x} - \frac{x}{3x - 9}\), we first factor the denominators:
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The first denominator \(x^2 - 3x\) can be factored as: \[ x^2 - 3x = x(x - 3) \]
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The second denominator \(3x - 9\) can be factored as: \[ 3x - 9 = 3(x - 3) \]
Now we can rewrite the expression: \[ \frac{3}{x(x - 3)} - \frac{x}{3(x - 3)} \]
Next, identify a common denominator, which will be \(3x(x - 3)\).
Now, rewrite each fraction with the common denominator:
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For the first term, multiply the numerator and denominator by 3: \[ \frac{3 \cdot 3}{x(x - 3) \cdot 3} = \frac{9}{3x(x - 3)} \]
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For the second term, multiply the numerator and denominator by \(x\): \[ \frac{x \cdot x}{3(x - 3) \cdot x} = \frac{x^2}{3x(x - 3)} \]
Now, the expression becomes: \[ \frac{9}{3x(x - 3)} - \frac{x^2}{3x(x - 3)} \]
Now we can combine the fractions: \[ \frac{9 - x^2}{3x(x - 3)} \]
The numerator \(9 - x^2\) can be rewritten as \(-(x^2 - 9)\), which factors further: \[ 9 - x^2 = -(x^2 - 9) = -(x - 3)(x + 3) \]
Thus, our combined expression is: \[ \frac{-(x - 3)(x + 3)}{3x(x - 3)} \]
Notice that \(x - 3\) cancels from the numerator and denominator (as long as \(x \neq 3\)), leading to: \[ \frac{-(x + 3)}{3x} \]
Thus, the simplified result for the original expression is: \[ -\frac{x + 3}{3x}, \quad x \neq 0, x \neq 3 \]
So the final answer is: \[ -\frac{x + 3}{3x} \]
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