Answers

2014-04-24T10:10:07+05:30
For x>1
 \frac{x}{x-1}+x=  \frac{x^2}{x-1}
 \frac{ x^{2}-x }{x-1}+x=0
 \frac{x(x-1)}{x-1}+x=0
 \frac{x^2-x+x^2-x}{y}= \frac{2x(x-1)}{x-1}=0
 so for x>1 no solution
for x<1
there is only one solution x=0
so relation holds only for x=0



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