Question

1+x^-1 over 1-x^-2 eliminate the complex fractions

Answer 1

they are kind of endless, huh?
\[\frac{1-x^{-1}}{1-x^{-2}}\]
\[\frac{1-x^{-1}}{1-x^{-2}}\times \frac{x^2}{x^2}=\frac{x^2-x}{x^2-1}\]

Answer 2

still not done though just like before

Answer 3

\[\frac{x^2-x}{x^2-1}=\frac{x(x-1)}{(x+1)(x-1)}\] then cancel to finish

Answer 4

Yea but is it the same because I put 1 + x^-1 and you put 1- x^-1

Answer 5

oops

Answer 6

change the minus to a plus and get
\[\frac{x^2+x}{x^2-1}=\frac{x(x+1)}{(x+1)(x-1)}=\frac{x}{x-1}\]

Answer 7

Do you happened to know x^-1+y^-1 over x+y

Answer 8

Begin with:
\[\frac{\frac{1}{x} + \frac{1}{y}}{x + y}\]
Re-write as:
\[\left(\frac{1}{x} + \frac{1}{y}\right) \div{(x + y)}\]
Reciprocate (x + y):
\[\left(\frac{1}{x} + \frac{1}{y}\right) \times \frac{1}{(x + y)}\]
Combine fractions 1/x and 1/y:
\[\left(\frac{y}{xy} + \frac{x}{xy}\right) \times \frac{1}{(x + y)}\]\[\left(\frac{x+y}{xy} \right) \times \frac{1}{(x + y)}\]
Multiply:
\[\frac{(x+y)}{xy(x+y)}\]

Cancel factors of 1:
\[\frac{\cancel{(x+y)}}{xy\cancel{(x+y)}}\]

Result:
\[\frac{1}{xy}\]