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Forkort om mulig brøkene
Posted: 03/05-2006 09:20
by Trond
[tex]\frac{{2x^2 + 2}}{{4x - 4}}[/tex]
[tex]\frac{{2x + 2}}{{x^2 + 2x + 1}}[/tex]
Re: Forkort om mulig brøkene
Posted: 03/05-2006 10:23
by ettam
[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]
Posted: 03/05-2006 10:27
by Trond
Fasiten viser [tex]\frac{{x^2 + 1}}{{2x - 2}}[/tex]
Re: Forkort om mulig brøkene
Posted: 03/05-2006 10:33
by Guest
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]