A) \[\frac{1}{{{x}^{2}}-1}\]
B) \[\frac{4x}{{{x}^{2}}-1}\]
C) \[\frac{x}{x-1}\]
D) \[\frac{x}{{{x}^{2}}-1}\]
Correct Answer: B
Solution :
\[\frac{x+1}{x-1}-\frac{1}{\frac{x+1}{x-1}}\] \[\frac{x+1}{x-1}-\frac{x-1}{x+1}\] Taking \[LCM=\frac{{{x}^{2}}+1+2x-{{x}^{2}}-1+2x}{(x-1)(x+1)}\Rightarrow \frac{4x}{{{x}^{2}}-1}\]
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