To simplify the expression \( x \left( 2^{-1}y + y^{-1}x^{-1} \right) \) without using negative indices and as a single fraction, we can follow these steps:
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Eliminate negative indices: Rewrite the negative indices with positive ones:
- \( 2^{-1} = \frac{1}{2} \)
- \( y^{-1} = \frac{1}{y} \)
- \( x^{-1} = \frac{1}{x} \)
So, we have: \[ x \left( \frac{1}{2}y + \frac{1}{y} \cdot \frac{1}{x} \right) = x \left( \frac{1}{2}y + \frac{1}{xy} \right) \]
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Combine the terms inside the parentheses: To combine \( \frac{1}{2}y \) and \( \frac{1}{xy} \), find a common denominator, which is \( 2y \): \[ \frac{1}{2}y = \frac{y^2}{2y} \quad \text{and} \quad \frac{1}{xy} = \frac{2}{2xy} \] Now, we can rewrite: \[ \frac{y^2}{2y} + \frac{2}{2xy} = \frac{y^2 + 2}{2y} \]
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Put it all together: Now replace the parentheses in the original expression: \[ x \left( \frac{y^2 + 2}{2y} \right) = \frac{x(y^2 + 2)}{2y} \]
Thus, the expression simplified into a single fraction is: \[ \frac{x(y^2 + 2)}{2y} \]