2.2 Övningar

(Skillnad mellan versioner)

Versionen från 28 april 2007 kl. 14.25

Lös ekvationerna

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

$\textrm{c) }$	$(x+3)^2-(x-5)^2=6x+4$	$\textrm{d) }$	$(x^2+4x+1)^2+3x^4-2x^2=(2x^2+2x+3)^2$

$\textrm{e) }$	$\displaystyle\frac{x+3}{x-3}-\displaystyle\frac{x+5}{x-2}=0$	$\textrm{f) }$	$\displaystyle\frac{4x}{4x-7}-\displaystyle\frac{1}{2x-3}=1$

Facit

Facit till alla delfrågorna

Lösning a

Lösning till delfråga a

Lösning b

Lösning till delfråga b

Lösning c

Lösning till delfråga c

Lösning d

Lösning till delfråga d

 ==Övning 2.2:1==
 <div class="ovning">
+L&ouml;s ekvationerna
+         <table width="100%" cellspacing="10px">
+         <tr align="left">
+           <td class="ntext">$\textrm{a) }$</td>
+           <td class="ntext">   $\displaystyle\frac{5x}{6}-\displaystyle\frac{x+2}{9}=\displaystyle\frac{1}{2}$</td>
+           <td class="ntext">$\textrm{b) }$</td>
+           <td class="ntext">$\displaystyle\frac{8x+3}{7}-\displaystyle\frac{5x-7}{4}=2$</td>
+         </tr>
+         <tr><td height="5px"/></tr>
+         <tr align="left">
+           <td class="ntext">$\textrm{c) }$</td>
+           <td class="ntext">$(x+3)^2-(x-5)^2=6x+4$</td>
+           <td class="ntext">$\textrm{d) }$</td>
+           <td class="ntext">$(x^2+4x+1)^2+3x^4-2x^2=(2x^2+2x+3)^2$</td>
+         </tr>
+         <tr><td height="5px"/></tr>
+         <tr align="left">
+           <td class="ntext">$\textrm{e) }$</td>
+           <td class="ntext">   $\displaystyle\frac{x+3}{x-3}-\displaystyle\frac{x+5}{x-2}=0$</td>
+           <td class="ntext">$\textrm{f) }$</td>
+           <td class="ntext">$\displaystyle\frac{4x}{4x-7}-\displaystyle\frac{1}{2x-3}=1$</td>
+         </tr>
+         <tr><td height="5px"/></tr>
+       </table>
 </div>