Solution 3.1:5b

From Förberedande kurs i matematik 1

(Difference between revisions)
Jump to: navigation, search
m (Lösning 3.1:5b moved to Solution 3.1:5b: Robot: moved page)
Line 1: Line 1:
-
{{NAVCONTENT_START}}
+
In order to eliminate
-
<center> [[Image:3_1_5b.gif]] </center>
+
<math>\sqrt[3]{7}=7^{{1}/{3}\;}</math>
-
{{NAVCONTENT_STOP}}
+
from the denominator, we can multiply the top and bottom of the fraction by
 +
<math>7^{{2}/{3}\;}</math>. The denominator becomes
 +
<math>7^{{1}/{3}\;}\centerdot 7^{{2}/{3}\;}=7^{{1}/{3+{2}/{3}\;}\;}=7^{1}=7</math>
 +
and we get
 +
 
 +
 
 +
<math>\frac{1}{\sqrt[3]{7}}=\frac{1}{7^{{1}/{3}\;}}=\frac{1}{7^{{1}/{3}\;}}\centerdot \frac{7^{{2}/{3}\;}}{7^{{2}/{3}\;}}=\frac{7^{{2}/{3}\;}}{7}.</math>

Revision as of 14:34, 22 September 2008

In order to eliminate \displaystyle \sqrt[3]{7}=7^{{1}/{3}\;} from the denominator, we can multiply the top and bottom of the fraction by \displaystyle 7^{{2}/{3}\;}. The denominator becomes \displaystyle 7^{{1}/{3}\;}\centerdot 7^{{2}/{3}\;}=7^{{1}/{3+{2}/{3}\;}\;}=7^{1}=7 and we get


\displaystyle \frac{1}{\sqrt[3]{7}}=\frac{1}{7^{{1}/{3}\;}}=\frac{1}{7^{{1}/{3}\;}}\centerdot \frac{7^{{2}/{3}\;}}{7^{{2}/{3}\;}}=\frac{7^{{2}/{3}\;}}{7}.

Retrieved from "http://wiki.sommarmatte.se/wikis/2008/bridgecourse1-ImperialCollege/index.php/Solution_3.1:5b"