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%%  Ein Beispiel der DANTE-Edition
%%  Mathematiksatz mit LaTeX
%%  3. Auflage
%%  Beispiel 09-21-1 auf Seite 203.
%%  Copyright (C) 2018 Herbert Voss
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\begin{document}
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\mdtheorem[style=theoremstyle]{definition}{Definition}
\begin{definition}[Quadratische Gleichung]  \addtolength\jot{5pt} \begin{align}
f(x)       &= g(x)\\ -x^{2}+6x-5 &= -\frac{1}{3}x^{2}+\frac{4}{3}x+\frac{5}{3}\\
         0 &= \frac{2}{3}x^{2}-\frac{14}{3}x+ \frac{20}{3}\label{eq:fg}\\
         0 &= x^{2}-7x+10\label{eq:4}\\  0 &= (x-2)(x-5)\\
\mathbb{L} &= \{2;\ 5\}\\[10pt]
F&=\int\limits_2^5\left(g(x)-f(x)\right)\diff x \\[10pt]
F &= \int\limits_2^5\left(\frac{2}{3}x^{2}\frac{14}{3}x+\frac{20}{3}\right)\diff x\\
  &= \left[\frac{2}{9}x^{3}-\frac{14}{6}x^{2}+\frac{20}{3}x\right]_{2}^{5}\\
  &= \underbrace{{\vphantom{\left(\frac{2}{9}\right)} \frac{2}{9}\cdot125-\frac{14}{6}\cdot25+\frac{20}{3}
     \cdot5}}_{\textrm{obere Grenze}}\underbrace{\vphantom{%
       \left(\frac{2}{9}\right)}-}_{-}\underbrace{\left(\frac{2}{9}
     \cdot8-\frac{14}{6}\cdot4+\frac{20}{3}\cdot 2\right)}_{\textrm{untere Grenze}}\\[10pt]
  &= \frac{500-1050+600}{18}-\frac{32-168+240}{18}\\
  &= \frac{50}{18}-\frac{104}{18}\\    &= -\frac{54}{18} = -3\\[10pt]
f(x) &= -x^{2}+6x-5=-\left(x^{2}-6x+5\right)\\   &= -\left(x-1\right)\left(x-5\right)\\[10pt]
f(x) &= -x^{2}+6x-5=-\left(x^{2}-6x+5\right)\\
     &= -\left(x^2-2\cdot3\cdot x+3^2-3^2+5\right)\\[-\normalbaselineskip]
     & \phantom{\mathrel{{}=-{}}(}\underbrace{\phantom{x^{2}-2\cdot3\cdot
        x+3^2}}_{\left(x-3\right)^2}\underbrace{\phantom{-3^2+5}}_{-4}\nonumber\\[10pt]
g(x) &= -\frac{1}{3}x^{2}+\frac{4}{3}x+\frac{5}{3}\\  &= -\frac{1}{3}\left(x^{2}-4x-5\right)\\
     &= -\frac{1}{3}\left(x+1\right)\left(x-5\right)\qquad\boxed{\mathbb{L}=\{-1;5\}}\\
g(x) &= -\frac{1}{3}x^{2}+\frac{4}{3}x+\frac{5}{3}\\ &= -\frac{1}{3}\left(x^{2}-4x-5\right)\\
     &= -\frac{1}{3}\left(x^{2}-2\cdot2\cdot x+2^{2}-2^{2}-5\right)\\
     &= -\frac{1}{3}\left(\left(x-2\right)^{2}-9\right)\\
     &= -\frac{1}{3}\left(x-2\right)^{2}+3\qquad\boxed{SP(2;3)}
\end{align} \end{definition}
\end{document}