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>