express as a single fraction

\(\dfrac{1}{x - 1} - \dfrac{2}{x} + \dfrac{1}{x + 1}\)

Guest Aug 21, 2022

#1**0 **

To express this as a single fraction, we consider the first two fractions:

\((\dfrac{1}{x-1}-\dfrac{2}{x})=\dfrac{x-2(x-1)}{x(x-1)}=\dfrac{x-2x+2}{x(x-1)}=\dfrac{2-x}{x(x-1)}\)

Then, we take this and combine it with the third fraction:

\((\dfrac{2-x}{x(x-1)}+\dfrac{1}{x+1})=\dfrac{(2-x)(x+1)+x(x-1)}{x(x-1)(x+1)}\)

Now, we simplify this:

\(\dfrac{2x+2-x^2-x+x^2-x}{x(x-1)(x+1)}=\dfrac{2}{x(x-1)(x+1)}\) which is our answer.

Another method is to make a common denominator:

Notice, the first term is missing x and (x+1), so we multiply top and bottom by x(x+1), and so on for the second, it is missing (x-1) and (x+1)

So we multiply top and bottom by (x-1)(x+1), and for the third, it is missing x and (x-1), so we multiply both top and bottom by x(x-1).

\(\dfrac{x(x+1)}{(x-1)x(x+1)}-\dfrac{2(x-1)(x+1)}{x(x-1)(x+1)}+\dfrac{x(x-1)}{(x+1)(x-1)x}\)

Now, since the denominators are the same, we can add the numerators:

\(\dfrac{x(x+1)-2(x-1)(x+1)+x(x-1)}{x(x-1)(x+1)}\)

And as before, we can simplify this by expanding the numerator.

I hope this helps.

Guest Aug 22, 2022