2.2 Übungen
Aus Online Mathematik Brückenkurs 1
(Unterschied zwischen Versionen)
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{| width="100%" cellspacing="10px" | {| width="100%" cellspacing="10px" | ||
|a) | |a) | ||
| - | || <math>\displaystyle\frac{x+3}{x-3}-\displaystyle\frac{x+5}{x-2}=0</math> | + | |width="100%" | <math>\displaystyle\frac{x+3}{x-3}-\displaystyle\frac{x+5}{x-2}=0</math> |
|- | |- | ||
|b) | |b) | ||
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|} | |} | ||
</div>{{#NAVCONTENT:Svar|Svar 2.2:3|Lösning a|Lösning 2.2:3a|Lösning b|Lösning 2.2:3b|Lösning c|Lösning 2.2:3c|Lösning d|Lösning 2.2:3d}} | </div>{{#NAVCONTENT:Svar|Svar 2.2:3|Lösning a|Lösning 2.2:3a|Lösning b|Lösning 2.2:3b|Lösning c|Lösning 2.2:3c|Lösning d|Lösning 2.2:3d}} | ||
| + | |||
| + | ===Övning 2.2:4=== | ||
| + | <div class="ovning"> | ||
| + | {| width="100%" cellspacing="10px" | ||
| + | |a) | ||
| + | |width="100%" | Skriv ekvationen för linjen<math>\,y=2x+3\,</math> på formen <math>\,y=kx+m\,</math> | ||
| + | |- | ||
| + | |b) | ||
| + | || Skriv ekvationen för linjen<math>,3x+4y-5=0</math> på formen <math>\,y=kx+m\,</math> | ||
| + | |- | ||
| + | |c) | ||
| + | || <math>\left(\displaystyle\frac{1}{x-1}-\frac{1}{x+1}\right)\left(x^2+\frac{1}{2}\right)=\displaystyle\frac{6x-1}{3x-3}</math> | ||
| + | |- | ||
| + | |d) | ||
| + | || <math>\left(\displaystyle\frac{2}{x}-3\right)\left(\displaystyle\frac{1}{4x}+\frac{1}{2}\right)-\left(\displaystyle\frac{1}{2x}-\frac{2}{3}\right)^2-\left(\displaystyle\frac{1}{2x}+\frac{1}{3}\right)\left(\displaystyle\frac{1}{2x}-\frac{1}{3}\right)=0</math> | ||
| + | |} | ||
| + | </div>{{#NAVCONTENT:Svar|Svar 2.2:4|Lösning a|Lösning 2.2:4a|Lösning b|Lösning 2.2:4b}} | ||
Version vom 13:04, 31. Mär. 2008
Övning 2.2:1
Lös ekvationerna
| a) | \displaystyle x-2=-1 | b) | \displaystyle 2x+1=13 |
| c) | \displaystyle \displaystyle\frac{1}{3}x-1=x | d) | \displaystyle 5x+7=2x-6 |
Svar
Laden...
Lösning a
Lösning b
Lösning c
Lösning d
Övning 2.2:2
Lös ekvationerna
| a) | \displaystyle \displaystyle\frac{5x}{6}-\displaystyle\frac{x+2}{9}=\displaystyle\frac{1}{2} | b) | \displaystyle \displaystyle\frac{8x+3}{7}-\displaystyle\frac{5x-7}{4}=2 |
| c) | \displaystyle (x+3)^2-(x-5)^2=6x+4 | d) | \displaystyle (x^2+4x+1)^2+3x^4-2x^2=(2x^2+2x+3)^2 |
Svar
Lösning a
Lösning b
Lösning c
Lösning d
Övning 2.2:3
Lös ekvationerna
| a) | \displaystyle \displaystyle\frac{x+3}{x-3}-\displaystyle\frac{x+5}{x-2}=0 |
| b) | \displaystyle \displaystyle\frac{4x}{4x-7}-\displaystyle\frac{1}{2x-3}=1 |
| c) | \displaystyle \left(\displaystyle\frac{1}{x-1}-\frac{1}{x+1}\right)\left(x^2+\frac{1}{2}\right)=\displaystyle\frac{6x-1}{3x-3} |
| d) | \displaystyle \left(\displaystyle\frac{2}{x}-3\right)\left(\displaystyle\frac{1}{4x}+\frac{1}{2}\right)-\left(\displaystyle\frac{1}{2x}-\frac{2}{3}\right)^2-\left(\displaystyle\frac{1}{2x}+\frac{1}{3}\right)\left(\displaystyle\frac{1}{2x}-\frac{1}{3}\right)=0 |
Svar
Lösning a
Lösning b
Lösning c
Lösning d
Övning 2.2:4
| a) | Skriv ekvationen för linjen\displaystyle \,y=2x+3\, på formen \displaystyle \,y=kx+m\, |
| b) | Skriv ekvationen för linjen\displaystyle ,3x+4y-5=0 på formen \displaystyle \,y=kx+m\, |
| c) | \displaystyle \left(\displaystyle\frac{1}{x-1}-\frac{1}{x+1}\right)\left(x^2+\frac{1}{2}\right)=\displaystyle\frac{6x-1}{3x-3} |
| d) | \displaystyle \left(\displaystyle\frac{2}{x}-3\right)\left(\displaystyle\frac{1}{4x}+\frac{1}{2}\right)-\left(\displaystyle\frac{1}{2x}-\frac{2}{3}\right)^2-\left(\displaystyle\frac{1}{2x}+\frac{1}{3}\right)\left(\displaystyle\frac{1}{2x}-\frac{1}{3}\right)=0 |
