[tex]\[\begin{array}{l}\frac{{6x^2 - 5x + 1}}{{2x^2 - x}}\\6x^2 - 5x + 1 = 0\\x = \frac{{ - \left( { - 5} \right) \pm \sqrt {\left( { - 5} \right)^2 - 4 \cdot 6 \cdot 1} }}{{2 \cdot 6}}\\x = \frac{{5 \pm \sqrt {25 - 24} }}{{12}}\\x = \frac{{5 \pm 1}}{{12}}\\x = \frac{6}{{12}}\,\,\,eller\,\,x = \frac{4}{{12}}\\x = \frac{1}{2}\,\,\,eller\,\,x = \frac{1}{3}\\\frac{{6\left( {x - \frac{1}{2}} \right)\left( {x - \frac{1}{3}} \right)}}{{2x^2 - x}} = \frac{{6\left( {x - \frac{1}{2}} \right)\left({x-\frac{1}{3}} \right)}}{{x\left( {2x - 1} \right)}} = \frac{{6\left( {2x-1} \right)\left( {3x - 1} \right)}}{{x\left( {2x - 1} \right)}}\\= \frac{{6\left( {2x - 1} \right)\left( {3x - 1} \right)}}{{x\left( {2x - 1} \right)}} = \frac{{6\left( {3x - 1} \right)}}{x}=\frac{{18x-1}}{x}\\\end{array} \][/tex]
Mens fasiten sier:
[tex]\[\frac{{3x - 1}}{x}\][/tex]
For meg virker det som de har glemt a i ax^2+b+c=a(x-x1)(x-x2)
Takk for hjelp
// alf
