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 \( x\left( 2^{-1}y + y^{-1}x^{-1} \right) \), we start by rewriting the negative indices:

1. \( 2^{-1} = \frac{1}{2} \)
2. \( y^{-1} = \frac{1}{y} \)
3. \( x^{-1} = \frac{1}{x} \)

Substituting these into the expression gives:

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

Now we can simplify \( \frac{1}{y} \cdot \frac{1}{x} = \frac{1}{xy} \). The expression now looks like this:

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

Next, we need to find a common denominator for the terms inside the parentheses. The common denominator of \( 2 \) and \( xy \) is \( 2xy \).

We can rewrite each term with the common denominator:

1. \( \frac{1}{2}y = \frac{y^2}{2y} \)
2. \( \frac{1}{xy} = \frac{2}{2xy} \)

Putting this together we have:

\[ \frac{y^2}{2} + \frac{2}{2xy} = \frac{y^2 \cdot x}{2xy} + \frac{2}{2xy} = \frac{y^2x + 2}{2xy} \]

Now substituting back into the expression, we get:

\[ x \left( \frac{y^2x + 2}{2xy} \right) = \frac{x(y^2x + 2)}{2xy} \]

The \( x \) in the numerator and denominator can be canceled (assuming \( x \neq 0 \)):

\[ \frac{y^2x + 2}{2y} \]

Thus, the simplified expression, as a single fraction, is:

\[ \frac{y^2x + 2}{2y} \]

Answer 2

Make it more simpler and use less brackets

Answer 3

Certainly! We can simplify the expression further and reduce the use of brackets.

Starting with the original expression:

\[ x \left( 2^{-1} y + y^{-1} x^{-1} \right) \]

We can rewrite the negative indices as:

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

This becomes:

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

Finding a common denominator for the terms \( \frac{1}{2} y \) and \( \frac{1}{xy} \) gives us \( 2xy \).

Rewriting each term with this common denominator, we get:

\[ \frac{y^2}{2} + \frac{2}{2xy} \]

This can be combined as:

\[ \frac{y^2 x + 2}{2xy} \]

When we multiply by \( x \):

\[ \frac{x(y^2 x + 2)}{2xy} \]

The \( x \) cancels out (assuming \( x \neq 0 \)), and we end up with:

\[ \frac{y^2 x + 2}{2y} \]

So the most simplified version in a single fraction with the minimum use of brackets is:

\[ \frac{y^2 x + 2}{2y} \]

This is the final, simplified form.