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{1}{x-x^2}-\displaystyle \frac{1}{x}</math>
+
|width="50%" | <math>x-2=-1</math>
|b)
|b)
-
|width="50%" | <math>\displaystyle \frac{1}{y^2-2y}-\displaystyle \frac{2}{y^2-4}</math>
+
|width="50%" | <math>2x+1=13</math>
|-
|-
|c)
|c)
-
|width="50%" | <math>\displaystyle \frac{(3x^2-12)(x^2-1)}{(x+1)(x+2)}</math>
+
|width="50%" | <math>\displaystyle\frac{1}{3}x-1=x</math>
|d)
|d)
-
|| <math>\displaystyle \frac{(y^2+4y+4)(2y-4)}{(y^2+4)(y^2-4)}</math>
+
|| <math>5x+7=2x-6</math>
|}
|}
</div>{{#NAVCONTENT:Svar|Svar 2.2:1|Lösning a|Lösning 2.2:1a|Lösning b|Lösning 2.2:1b|Lösning c|Lösning 2.2:1c|Lösning d|Lösning 2.2:1d}}
</div>{{#NAVCONTENT:Svar|Svar 2.2:1|Lösning a|Lösning 2.2:1a|Lösning b|Lösning 2.2:1b|Lösning c|Lösning 2.2:1c|Lösning d|Lösning 2.2:1d}}

Revision as of 12:39, 31 March 2008

 

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Övning 2.2:1

Förenkla så långt som möjligt

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