Bunuel wrote:
(\frac{x+1}{x-1})^2
If x#0 and x#1, and if x is replaced by 1/x everywhere in the expression above, then the resulting expression is equivalent to
A. (\frac{x+1}{x-1})^2
B. (\frac{x-1}{x+1})^2
C. \frac{x^2+1}{1-x^2}
D. \frac{x^2-1}{x^2+1}
E. -(\frac{x-1}{x+1})^2
(\frac{\frac{1}{x}+1}{\frac{1}{x}-1})^2=(\frac{\frac{1+x}{x}}{\frac{1-x}{x}})^2=(\frac{1+x}{1-x})^2=(\frac{x+1}{x-1})^2.
Answer: A.
If x#0 and x#1, and if x is replaced by 1/x everywhere in the expression above, then the resulting expression is equivalent to
A. (\frac{x+1}{x-1})^2
B. (\frac{x-1}{x+1})^2
C. \frac{x^2+1}{1-x^2}
D. \frac{x^2-1}{x^2+1}
E. -(\frac{x-1}{x+1})^2
(\frac{\frac{1}{x}+1}{\frac{1}{x}-1})^2=(\frac{\frac{1+x}{x}}{\frac{1-x}{x}})^2=(\frac{1+x}{1-x})^2=(\frac{x+1}{x-1})^2.
Answer: A.
I dont get (\frac{\frac{1}{x}+1}{\frac{1}{x}-1})^2=(\frac{\frac{1+x}{x}}{\frac{1-x}{x}})^2.. could you please explain your steps in a few words?
