Solution 3.1:5b

From Förberedande kurs i matematik 1

(Difference between revisions)
Jump to: navigation, search
Current revision (11:22, 30 September 2008) (edit) (undo)
m
 
Line 1: Line 1:
-
In order to eliminate
+
In order to eliminate <math>\sqrt[3]{7} = 7^{1/3}</math> 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}\cdot 7^{2/3} = 7^{1/3+2/3} = 7^1 = 7</math> and we get
-
<math>\sqrt[3]{7}=7^{{1}/{3}\;}</math>
+
-
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
+
-
 
+
{{Displayed math||<math>\frac{1}{\sqrt[3]{7}} = \frac{1}{7^{1/3}} = \frac{1}{7^{1/3}}\cdot \frac{7^{2/3}}{7^{2/3}} = \frac{7^{2/3}}{7}\,\textrm{.}</math>}}
-
<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>
+

Current revision

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}\cdot 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}}\cdot \frac{7^{2/3}}{7^{2/3}} = \frac{7^{2/3}}{7}\,\textrm{.}
Retrieved from "http://wiki.sommarmatte.se/wikis/2008/bridgecourse1-ImperialCollege/index.php/Solution_3.1:5b"