Forkort om mulig brøkene

Trond · 03/05-2006 09:20

[tex]\frac{{2x^2 + 2}}{{4x - 4}}[/tex]

[tex]\frac{{2x + 2}}{{x^2 + 2x + 1}}[/tex]

ettam · 03/05-2006 10:23

[tex]\frac{{2x^2 + 2}}{{4x - 4}} = \frac{{2(x^2 + 2)}}{{2(2x - 2)}} = \frac{{x^2 + 2}}{{2x - 2}}[/tex]

[tex]\frac{{2x + 2}}{{x^2 + 2x + 1}} = \frac{{2(x + 1)}}{{(x + 1)^2 }} = \underline{\underline {\frac{2}{{x + 1}}}}[/tex]

Trond · 03/05-2006 10:27

Fasiten viser [tex]\frac{{x^2 + 1}}{{2x - 2}}[/tex]

Guest · 03/05-2006 10:33

ettam wrote:[tex]\frac{{2x^2 + 2}}{{4x - 4}} = \frac{{2(x^2 + 2)}}{{2(2x - 2)}} = \frac{{x^2 + 2}}{{2x - 2}}[/tex]

[tex]\frac{{2x + 2}}{{x^2 + 2x + 1}} = \frac{{2(x + 1)}}{{(x + 1)^2 }} = \underline{\underline {\frac{2}{{x + 1}}}}[/tex]

Den første brøken skal være:
[tex]\frac{{2x^2 + 2}}{{4x - 4}} = \frac{{2(x^2 + 1)}}{{2(2x - 2)}} = \frac{{x^2 + 1}}{{2x - 2}}[/tex]