2.2 Övningar

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Versionen från 28 april 2007 kl. 15.45 (redigera)
Ossiang (Diskussion | bidrag)
(Övning 2.2:3)
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Versionen från 28 april 2007 kl. 15.47 (redigera) (ogör)
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(Övning 2.2:3)
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Rad 155: Rad 155:
<td class="ntext">a)</td> <td class="ntext">a)</td>
<td class="ntext">$\displaystyle\frac{x+3}{x-3}-\displaystyle\frac{x+5}{x-2}=0$</td> <td class="ntext">$\displaystyle\frac{x+3}{x-3}-\displaystyle\frac{x+5}{x-2}=0$</td>
 +</tr><tr align="left">
<td class="ntext">b)</td> <td class="ntext">b)</td>
<td class="ntext">$\displaystyle\frac{4x}{4x-7}-\displaystyle\frac{1}{2x-3}=1$</td> <td class="ntext">$\displaystyle\frac{4x}{4x-7}-\displaystyle\frac{1}{2x-3}=1$</td>
Rad 162: Rad 163:
<td class="ntext">c)</td> <td class="ntext">c)</td>
<td class="ntext">$(\displaystyle\frac{1}{x-1}-\frac{1}{x+1})(x^2+\frac{1}{2})=\frac{6x-1}{3x-3}$</td> <td class="ntext">$(\displaystyle\frac{1}{x-1}-\frac{1}{x+1})(x^2+\frac{1}{2})=\frac{6x-1}{3x-3}$</td>
-<td class="ntext">d) </td>+<td class="ntext">d) </td></tr><tr align="left">
-<td class="ntext">$\left(\displaystyle\frac{2}{x}-3\right)\left(\displaystyle\frac{1}{4}+\frac{1}{2}\right)-\left(\displaystyle\frac{1}{2x}-\frac{2}{3}\right)^2-\left(\displaystyle\frac{1}{2x}+\frac{1}{3}\right)\left(\displaystyle\frac{1}{2x}-\frac{1}{3}\right)$</td>+<td class="ntext">$\left(\displaystyle\frac{2}{x}-3\right)\left(\displaystyle\frac{1}{4}+\frac{1}{2}\right)-\left(\displaystyle\frac{1}{2x}-\frac{2}{3}\right)^2-\left(\displaystyle\frac{1}{2x}+\frac{1}{3}\right)\left(\displaystyle\frac{1}{2x}-\frac{1}{3}\right)=0$</td>
</tr> </tr>
<tr><td height="5px"/></tr> <tr><td height="5px"/></tr>

Versionen från 28 april 2007 kl. 15.47

Övning 2.2:1

Lös ekvationerna

a) $x-2=-1$ b) $2x+1=13$
c) $\frac{1}{3}x-1=x$ d) $5x+7=2x-6$

Övning 2.2:2

Lös ekvationerna

a) $\displaystyle\frac{5x}{6}-\displaystyle\frac{x+2}{9}=\displaystyle\frac{1}{2}$ b) $\displaystyle\frac{8x+3}{7}-\displaystyle\frac{5x-7}{4}=2$
c) $(x+3)^2-(x-5)^2=6x+4$ d) $(x^2+4x+1)^2+3x^4-2x^2=(2x^2+2x+3)^2$

Övning 2.2:3

Lös ekvationerna

a) $\displaystyle\frac{x+3}{x-3}-\displaystyle\frac{x+5}{x-2}=0$
b) $\displaystyle\frac{4x}{4x-7}-\displaystyle\frac{1}{2x-3}=1$
c) $(\displaystyle\frac{1}{x-1}-\frac{1}{x+1})(x^2+\frac{1}{2})=\frac{6x-1}{3x-3}$ d)
$\left(\displaystyle\frac{2}{x}-3\right)\left(\displaystyle\frac{1}{4}+\frac{1}{2}\right)-\left(\displaystyle\frac{1}{2x}-\frac{2}{3}\right)^2-\left(\displaystyle\frac{1}{2x}+\frac{1}{3}\right)\left(\displaystyle\frac{1}{2x}-\frac{1}{3}\right)=0$
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