If \(x + \frac{1}{x} = 5\) , then find the value of \(\frac{6x}{ x^{2} + x +1} \)
1
Given that \(x + \frac{1}{x} = 5\), we want to find the value of \(\frac{6x}{x^2 + x + 1}\).
We can rewrite the expression as follows:
\(\frac{6x}{x^2 + x + 1} = \frac{6x}{x^2 + x + 1} \times \frac{\frac{1}{x}}{\frac{1}{x}} = \frac{6}{x + 1 + \frac{1}{x}}\)
Since \(x + \frac{1}{x} = 5\), we can substitute this into the expression:
\(\frac{6}{x + 1 + \frac{1}{x}} = \frac{6}{5 + 1} = \frac{6}{6} = 1\)
Therefore, the value of \(\frac{6x}{x^2 + x + 1}\) is 1.
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