| If \[x+\frac{1}{x}=9,\]\[(x\ne 0),\] then the value of \[\frac{{{x}^{2}}-4x+1}{{{x}^{2}}+6x+1}\] is |
A) \[\frac{1}{3}\]
B) \[\frac{2}{3}\]
C) 3
D) 4
Correct Answer: A
Solution :
| \[\frac{{{x}^{2}}-4x+1}{{{x}^{2}}+6x+1}\,\,=\,\,\frac{x-4+\frac{1}{x}}{x+6+\frac{1}{x}}\,\,=\,\,\frac{x+\frac{1}{x}-4}{x+\frac{1}{x}+6}\] |
| \[=\frac{9-4}{9+6}\,\,=\,\,\frac{5}{15}=\frac{1}{3}\] |
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