$$\begin{array}{rcl} \frac{2}{x-1}+\frac{1}{x-2}& = & \frac{x}{\left(x-1\right)\left(x-2\right)}\\ & & \\ \frac{2·\cancel{\left(x-1\right)}\left(x-2\right)}{\cancel{x-1}}+\frac{1·\left(x-1\right)\cancel{\left(x-2\right)}}{\cancel{x-2}}& =& \frac{x·\cancel{\left(x-1\right)\left(x-2\right)}}{\cancel{\left(x-1\right)\left(x-2\right)}}\\ & & \\ 2\left(x-2\right)+\left(x-1\right)& =& x\\ & & \\ 2x-4+x-1& =& x\\ & & \\ 2x+x-x& =& 4+1\\ & & \\ 2x& =& 5\\ & & \\ x& =& \frac{5}{2}\end{array}$$

Verdien av $$x$$ gir ikke null i nevner. Løsning er derfor gyldig.

$$\begin{array}{rcl} \frac{2}{2\left(x-1\right)}+\frac{1}{x-2}& = & \frac{x-3}{\left(x-1\right)\left(x-2\right)}\\ & & \\ \frac{2·\cancel{2}\cancel{\left(x-1\right)}\left(x-2\right)}{\cancel{2}\cancel{\left(x-1\right)}}+\frac{1·2\left(x-1\right)\cancel{\left(x-2\right)}}{\cancel{x-2}}& =& \frac{\left(x-3\right)·2\cancel{\left(x-1\right)\left(x-2\right)}}{\cancel{\left(x-1\right)\left(x-2\right)}}\\ & & \\ 2\left(x-2\right)+2\left(x-1\right)& =& 2\left(x-3\right)\\ & & \\ 2x-4+2x-2& =& 2x-6\\ & & \\ 2 x+2x-2x& =& 4+2-6\\ & & \\ 2x& =& 0\\ & & \\ x& =& 0\end{array}$$

Verdien av $$x$$ gir ikke null i nevner. Løsning er derfor gyldig.

$$\begin{array}{rcl} \frac{3}{2\left(x-1\right)}-\frac{1}{x-2}& = & \frac{x-3}{\left(x-1\right)\left(x-2\right)}\\ & & \\ \frac{3·\cancel{2}\cancel{\left(x-1\right)}\left(x-2\right)}{\cancel{2}\cancel{\left(x-1\right)}}-\frac{1·2\left(x-1\right)\cancel{\left(x-2\right)}}{\cancel{x-2}}& =& \frac{\left(x-3\right)·2\cancel{\left(x-1\right)\left(x-2\right)}}{\cancel{\left(x-1\right)\left(x-2\right)}}\\ & & \\ 3\left(x-2\right)-2\left(x-1\right)& =& 2\left(x-3\right)\\ & & \\ 3x-6-2x+2& =& 2x-6\\ & & \\ 3x-2x-2x& =& 6-2-6\\ & & \\ -x& =& -2\\ & & \\ x& =& 2\end{array}$$

Her fikk vi som eneste løsning en verdi for $$x$$ som gir null i nevner. Løsningen kan derfor ikke aksepteres, og likningen har ingen løsning.

Merk hvordan GeoGebra markerer at likningen ikke har løsning.