2.1 Exercises

From Förberedande kurs i matematik 1

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(Ny sida: __NOTOC__ {| border="0" cellspacing="0" cellpadding="0" height="30" width="100%" | style="border-bottom:1px solid #000" width="5px" |   {{Mall:Ej vald flik|Teori}...)
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|width="33%" | <math>(1+x-x^2)xy</math>
|width="33%" | <math>(1+x-x^2)xy</math>
|c)
|c)
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|width="33%" | <math>\displaystyle \frac{2}{7}-\frac{1}{5}</math>
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|width="33%" | <math>-x^2(4-y^2)</math>
|-
|-
|d)
|d)
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|| <math>\displaystyle \frac{1}{3}+\frac{1}{4}+\frac{1}{5}</math>
+
|| <math>x^3y^2\left(\displaystyle \frac{1}{y} - \frac{1}{xy}+1\right)</math>
|e)
|e)
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||<math> \displaystyle \frac{8}{7}+\frac{3}{4}-\frac{4}{3}</math>
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||<math> (x-7)^2</math>
|f)
|f)
-
||<math> \displaystyle \frac{8}{7}+\frac{3}{4}-\frac{4}{3}</math>
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||<math> (5+4y)^2</math>
|-
|-
|g)
|g)
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||<math> \displaystyle \frac{8}{7}+\frac{3}{4}-\frac{4}{3}</math>
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||<math>(y^2-3x^3)^2</math>
|h)
|h)
-
||<math> \displaystyle \frac{8}{7}+\frac{3}{4}-\frac{4}{3}</math>
+
||<math> (5x^3+3x^5)^2</math>
|}
|}
</div>{{#NAVCONTENT:Svar|Svar 2.1:1|Lösning a|Lösning 2.1:1a|Lösning b|Lösning 2.1:1b|Lösning c|Lösning 2.1:1c|Lösning d|Lösning 2.1:1d|Lösning e|Lösning 2.1:1e|Lösning f|Lösning 2.1:1f|Lösning g|Lösning 2.1:1g|Lösning h|Lösning 2.1:1h}}
</div>{{#NAVCONTENT:Svar|Svar 2.1:1|Lösning a|Lösning 2.1:1a|Lösning b|Lösning 2.1:1b|Lösning c|Lösning 2.1:1c|Lösning d|Lösning 2.1:1d|Lösning e|Lösning 2.1:1e|Lösning f|Lösning 2.1:1f|Lösning g|Lösning 2.1:1g|Lösning h|Lösning 2.1:1h}}

Revision as of 09:17, 28 March 2008

 

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Övning 2.1:1

Skriv på gemensamt bråkstreck

a) \displaystyle 3x(x-1) b) \displaystyle (1+x-x^2)xy c) \displaystyle -x^2(4-y^2)
d) \displaystyle x^3y^2\left(\displaystyle \frac{1}{y} - \frac{1}{xy}+1\right) e) \displaystyle (x-7)^2 f) \displaystyle (5+4y)^2
g) \displaystyle (y^2-3x^3)^2 h) \displaystyle (5x^3+3x^5)^2