Let \(\alpha\) and \(\beta\) be the roots of the equation \(x^2 - p(x+1) - q = 0\). What is \(\dfrac{\alpha^2+2\alpha+1}{\alpha^2+2\alpha+q} + \dfrac{\beta^2+2\beta+1}{\beta^2+2\beta+q}\) equal to?
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Since \(\alpha, \beta\) satisfy \(x^2 = p(x+1)+q\), setting \(u=\alpha+1, v=\beta+1\) gives \(u^2-(p+2)u+(1-q)=0\) (and similarly for \(v\)), so \(u+v=p+2\) and \(uv=1-q\). The given expression equals \(\dfrac{u^2}{u^2+q-1}+\dfrac{v^2}{v^2+q-1} = \dfrac{u^2}{u(u-v)} - \dfrac{v^2}{v(u-v)} = \dfrac{u-v}{u-v} = 1\).
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