{Najdi konstanty $a$ a $b$, je-li $$
   \lix\left(\frac{x^2+1}{x+1}-ax-b\right)=0.
$$}{
\begin{eqnarray*}
   \lix\left(\frac{x^2+1}{x+1}  - ax - b\right) & = & 0\\
   \lix\left(\frac{x^2-1+2}{x+1}- ax - b\right) & = & 0\\
   \lix\left(x-1+\frac2{x+1}    - ax - b\right) & = & 0\\
   \lix\frac2{x+1}                              & = & 0\\
                                          & \Downarrow &  \\
                                            x-1 & = & ax+b\\
                                          & \Downarrow &  \\
   a=1 & \wedge & b=-1.
\end{eqnarray*}
}
