Lösung 3.1:5c
Aus Online Mathematik Brückenkurs 1
(Unterschied zwischen Versionen)
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| - | The trick is to use the | + | The trick is to use the formula for the difference of two squares |
| - | <math> | + | <math>(a-b)(a+b) = a^{2}-b^{2}</math> and multiply the top and bottom of the fraction by <math>3-\sqrt{7}</math> (note the minus sign), since then the new denominator will be <math>(3+\sqrt{7})(3-\sqrt{7}) = 3^{2} - (\sqrt{7})^{2} = 9-7 = 2</math> (the formula with <math>a=3</math> and <math>b=\sqrt{7}\,</math>), i.e. the root sign is squared away. |
| - | and multiply the top and bottom of the fraction by | + | |
| - | <math>3-\sqrt{7}</math> | + | |
| - | (note the minus sign), since then the new denominator will be | + | |
| - | <math> | + | |
| - | ( | + | |
| - | <math>a= | + | |
| - | and | + | |
| - | <math>b=\sqrt | + | |
| - | ), i.e. the root sign is squared away. | + | |
The whole calculation is | The whole calculation is | ||
| - | + | {{Displayed math||<math>\begin{align} | |
| - | <math>\begin{align} | + | \frac{2}{3+\sqrt{7}} |
| - | + | &= \frac{2}{3+\sqrt{7}}\cdot\frac{3-\sqrt{7}}{3-\sqrt{7}} | |
| - | & =\frac{2\ | + | = \frac{2(3-\sqrt{7}\,)}{3^{2}-(\sqrt{7}\,)^{2}}\\[5pt] |
| - | \end{align}</math> | + | &= \frac{2\cdot 3-2\sqrt{7}}{2} = 3-\sqrt{7}\,\textrm{.} |
| + | \end{align}</math>}} | ||
Version vom 11:29, 30. Sep. 2008
The trick is to use the formula for the difference of two squares \displaystyle (a-b)(a+b) = a^{2}-b^{2} and multiply the top and bottom of the fraction by \displaystyle 3-\sqrt{7} (note the minus sign), since then the new denominator will be \displaystyle (3+\sqrt{7})(3-\sqrt{7}) = 3^{2} - (\sqrt{7})^{2} = 9-7 = 2 (the formula with \displaystyle a=3 and \displaystyle b=\sqrt{7}\,), i.e. the root sign is squared away.
The whole calculation is
