If x = 2 + \(\sqrt {3}\)

Find \(\frac{x^2 - x + 1}{x^2 + 1 + x}\)

\(\frac{3}{5}\)

Formula → If x = a + b and a² - b² = 1, then \(\frac{1}{x}\) = a - b always

⇒ x = 2 + \(\sqrt {3}\)

⇒ \(\frac{1}{x}\) = 2 - \(\sqrt {3}\)

Now put the values and find,

\(\frac{x + \frac{1}{x} - 1}{x + \frac{1}{x} + 1}\) = \(\frac{2 + \sqrt {3} + 2 - \sqrt {3} - 1}{2 + \sqrt {3} + 2 - \sqrt {3} + 1}\) = \(\frac{3}{5}\)