Hvordan få samme brøk?
Moderatorer: Aleks855, Gustav, Nebuchadnezzar, Janhaa, DennisChristensen, Emilga
[tex]1\cdot \sqrt{2x-1}+x\cdot \frac{1}{\cancel{2}\sqrt{2x-1}} \cdot\cancel{2}[/tex]
[tex]\sqrt{2x-1}+\frac{x}{\sqrt{2x-1}}[/tex] Fellesnevner er $\sqrt{2x-1}$
[tex]\sqrt{2x-1} \cdot \frac{\sqrt{2x-1}}{\sqrt{2x-1}}+\frac{x}{\sqrt{2x-1}}[/tex] (husk at $\sqrt{a} \cdot \sqrt{a}=a$)
[tex]\frac{2x-1}{\sqrt{2x-1}}+\frac{x}{\sqrt{2x-1}}[/tex]
[tex]\sqrt{2x-1}+\frac{x}{\sqrt{2x-1}}[/tex] Fellesnevner er $\sqrt{2x-1}$
[tex]\sqrt{2x-1} \cdot \frac{\sqrt{2x-1}}{\sqrt{2x-1}}+\frac{x}{\sqrt{2x-1}}[/tex] (husk at $\sqrt{a} \cdot \sqrt{a}=a$)
[tex]\frac{2x-1}{\sqrt{2x-1}}+\frac{x}{\sqrt{2x-1}}[/tex]
Ah, jeg hadde glemt (husk at $\sqrt{a} \cdot \sqrt{a}=a$)Kjemikern skrev:[tex]1\cdot \sqrt{2x-1}+x\cdot \frac{1}{\cancel{2}\sqrt{2x-1}} \cdot\cancel{2}[/tex]
[tex]\sqrt{2x-1}+\frac{x}{\sqrt{2x-1}}[/tex] Fellesnevner er $\sqrt{2x-1}$
[tex]\sqrt{2x-1} \cdot \frac{\sqrt{2x-1}}{\sqrt{2x-1}}+\frac{x}{\sqrt{2x-1}}[/tex] (husk at $\sqrt{a} \cdot \sqrt{a}=a$)
[tex]\frac{2x-1}{\sqrt{2x-1}}+\frac{x}{\sqrt{2x-1}}[/tex]
Tusen takk!