2.2 Exercises

From Förberedande kurs i matematik 1

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{| width="100%" cellspacing="10px"
{| width="100%" cellspacing="10px"
|a)
|a)
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|width="50%" | <math>\displaystyle\frac{5x}{6}-\displaystyle\frac{x+2}{9}=\displaystyle\frac{1}{2}</math>
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|width="50%" | <math>\displaystyle\frac{5x}{6}-\displaystyle\frac{x+2}{9}=\displaystyle\frac{1}{2}</math>
|b)
|b)
|width="50%" | <math>\displaystyle\frac{8x+3}{7}-\displaystyle\frac{5x-7}{4}=2</math>
|width="50%" | <math>\displaystyle\frac{8x+3}{7}-\displaystyle\frac{5x-7}{4}=2</math>
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</div>{{#NAVCONTENT:Svar|Svar 2.2:2|Lösning a|Lösning 2.2:2a|Lösning b|Lösning 2.2:2b|Lösning c|Lösning 2.2:2c|Lösning d|Lösning 2.2:2d}}
</div>{{#NAVCONTENT:Svar|Svar 2.2:2|Lösning a|Lösning 2.2:2a|Lösning b|Lösning 2.2:2b|Lösning c|Lösning 2.2:2c|Lösning d|Lösning 2.2:2d}}
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===Övning 2.2:3===
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<div class="ovning">
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L&ouml;s ekvationerna
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{| width="100%" cellspacing="10px"
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|a)
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|| <math>\displaystyle\frac{x+3}{x-3}-\displaystyle\frac{x+5}{x-2}=0</math>
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|-
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|b)
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|| <math>\displaystyle\frac{4x}{4x-7}-\displaystyle\frac{1}{2x-3}=1</math>
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|-
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|c)
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|| <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>
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|-
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|d)
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|| <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>
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|}
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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}}

Revision as of 12:55, 31 March 2008

 

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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

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

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