$p^{2} - p + \frac{1}{4} \div p - \frac{1}{2}$

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Solution

$p^{2} - p + \frac{1}{4} = p^{2} - 2 (p) (\frac{1}{2}) + (\frac{1}{2})^{2} = (p - \frac{1}{2})^{2}$
$p^{2} - p + \frac{1}{4} \div p - \frac{1}{2} = \frac{(p - \frac{1}{2})^{2}}{p - \frac{1}{2}} = p - \frac{1}{2}$

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P + \frac{2}{1 + \frac{1}{P + \frac{1}{1 + P . . .}}} = \sqrt{P^{2} + 4 P}

(p^{2} + 5 p + 4) \div (p + 1)

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4

P (2) = - 1, P^{'} (2) = 0, P^{''} (2) = 2, P^{'''} (2) = - 12, P^{''''} (2) = 24

P^{^{''}} (1) =

p = 5 - 2 \sqrt{6},

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P^{'''} (2) = - 12

P^{i v} (2) = 24

P^{''} (1)

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