Write this as simply as possible without using negative indices and as a single fraction:

Question

x [(2^(-1)y + y^(-1)x^(-1)].

Answer 1

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:

1. 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: \[ x \left( \frac{1}{2}y + \frac{1}{y} \cdot \frac{1}{x} \right) = x \left( \frac{1}{2}y + \frac{1}{xy} \right) \]

2. 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: \[ \frac{y^2}{2y} + \frac{2}{2xy} = \frac{y^2 + 2}{2y} \]

3. 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: \[ \frac{x(y^2 + 2)}{2y} \]

Answer 2

Can your be more simpler with less brackets

Answer 3

Certainly! Let's simplify it even more with minimal brackets and present it cleanly.

Starting from the original expression: \[ x \left( 2^{-1}y + y^{-1}x^{-1} \right) \]

We rewrite it without negative indices: \[ x \left( \frac{1}{2}y + \frac{1}{y} \cdot \frac{1}{x} \right) = x \left( \frac{1}{2}y + \frac{1}{xy} \right) \]

Next, we use a common denominator: The common denominator will be \( 2y \): \[ \frac{1}{2}y = \frac{y^2}{2y} \quad \text{and} \quad \frac{1}{xy} = \frac{2}{2xy} \]

Now we can add these two fractions: \[ \frac{y^2 + 2}{2y} \]

Finally, multiply by \( x \): \[ x \cdot \frac{y^2 + 2}{2y} = \frac{x(y^2 + 2)}{2y} \]

So the final simplified expression is: \[ \frac{x(y^2 + 2)}{2y} \]

This is as simple and direct as it gets with minimal brackets.