\ANG{
Berechne die L"osungsmenge:\qquad
$\D\sqrt{x+1}=\sqrt{x}+\sqrt{x-1}$
}
\LSG{
$x>1\quad\DANN\quad\DD=[1,+\infty[$
\begin{align}
  x+1&=x+x-1+2\sqrt{x^{2}-x}\nonumber\\
  2-x&=2\sqrt{x^{2}-x}\nonumber\\
  4-4x+x^{2}&=4x^{2}-4x\nonumber\\
  x^{2}&=\frac43\quad\DANN\quad x_{1}=\frac23\sqrt{3},\quad
  \left(x_{2}=-\frac23\sqrt{3}\notin\DD\right)\nonumber
\end{align}
}
