Asad wrote:
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 equivalentto
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 equivalentto
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.
Bunuel
If we put x=1 we get infinite result. So, x≠1. But what's the problem when x=0? I mean why condition says that x≠0, too? If x=0, what this will effect in the
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