Lösung 3.1:3b
Aus Online Mathematik Brückenkurs 1
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By successively dividing by 2 and 3, we see that | By successively dividing by 2 and 3, we see that | ||
| - | {{ | + | {{Abgesetzte Formel||<math>\begin{align} |
96 &= 2\cdot 48 = 2\cdot 2\cdot 24 = 2\cdot 2\cdot 2\cdot 12 = 2\cdot 2\cdot 2\cdot 2\cdot 6\\ | 96 &= 2\cdot 48 = 2\cdot 2\cdot 24 = 2\cdot 2\cdot 2\cdot 12 = 2\cdot 2\cdot 2\cdot 2\cdot 6\\ | ||
&= 2\cdot 2\cdot 2\cdot 2\cdot 2\cdot 3 = 2^{5}\cdot 3,\\[5pt] | &= 2\cdot 2\cdot 2\cdot 2\cdot 2\cdot 3 = 2^{5}\cdot 3,\\[5pt] | ||
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Thus, | Thus, | ||
| - | {{ | + | {{Abgesetzte Formel||<math>\begin{align} |
\sqrt{96} &= \sqrt{2^{5}\cdot 3} = \sqrt{2^{2}\cdot 2^{2}\cdot 2\cdot 3} = 2\cdot 2\cdot \sqrt{2}\cdot \sqrt{3}\,,\\[5pt] | \sqrt{96} &= \sqrt{2^{5}\cdot 3} = \sqrt{2^{2}\cdot 2^{2}\cdot 2\cdot 3} = 2\cdot 2\cdot \sqrt{2}\cdot \sqrt{3}\,,\\[5pt] | ||
\sqrt{18} &= \sqrt{2\cdot 3^{2}} = 3\cdot\sqrt{2}\,, | \sqrt{18} &= \sqrt{2\cdot 3^{2}} = 3\cdot\sqrt{2}\,, | ||
| Zeile 18: | Zeile 18: | ||
and the whole quotient can be written as | and the whole quotient can be written as | ||
| - | {{ | + | {{Abgesetzte Formel||<math>\frac{\sqrt{96}}{\sqrt{18}} = \frac{2\cdot 2\cdot \sqrt{2}\cdot \sqrt{3}}{3\cdot \sqrt{2}} = \frac{4\sqrt{3}}{3}\,\textrm{.}</math>}} |
Note: If it is difficult to work with radicals, it is possible instead to write everything in power form | Note: If it is difficult to work with radicals, it is possible instead to write everything in power form | ||
| - | {{ | + | {{Abgesetzte Formel||<math>\begin{align} |
\frac{\sqrt{96}}{\sqrt{18}} | \frac{\sqrt{96}}{\sqrt{18}} | ||
&= \frac{96^{1/2}}{18^{1/2}} | &= \frac{96^{1/2}}{18^{1/2}} | ||
Version vom 08:36, 22. Okt. 2008
When simplifying a radical expression, a common technique is to divide up the numbers under the root sign into their smallest possible integer factors and then take out the squares and see if common factors cancel each other out or can be combined together in a new way.
By successively dividing by 2 and 3, we see that
| \displaystyle \begin{align}
96 &= 2\cdot 48 = 2\cdot 2\cdot 24 = 2\cdot 2\cdot 2\cdot 12 = 2\cdot 2\cdot 2\cdot 2\cdot 6\\ &= 2\cdot 2\cdot 2\cdot 2\cdot 2\cdot 3 = 2^{5}\cdot 3,\\[5pt] 18 &= 2\cdot 9 = 2\cdot 3\cdot 3 = 2\cdot 3^{2}. \end{align} |
Thus,
| \displaystyle \begin{align}
\sqrt{96} &= \sqrt{2^{5}\cdot 3} = \sqrt{2^{2}\cdot 2^{2}\cdot 2\cdot 3} = 2\cdot 2\cdot \sqrt{2}\cdot \sqrt{3}\,,\\[5pt] \sqrt{18} &= \sqrt{2\cdot 3^{2}} = 3\cdot\sqrt{2}\,, \end{align} |
and the whole quotient can be written as
| \displaystyle \frac{\sqrt{96}}{\sqrt{18}} = \frac{2\cdot 2\cdot \sqrt{2}\cdot \sqrt{3}}{3\cdot \sqrt{2}} = \frac{4\sqrt{3}}{3}\,\textrm{.} |
Note: If it is difficult to work with radicals, it is possible instead to write everything in power form
| \displaystyle \begin{align}
\frac{\sqrt{96}}{\sqrt{18}} &= \frac{96^{1/2}}{18^{1/2}} = \frac{(2^{5}\cdot 3)^{1/2}}{(2\cdot 3^{2})^{1/2}} = \frac{2^{5\cdot\frac{1}{2}}\cdot 3^{\frac{1}{2}}}{2^{\frac{1}{2}}\cdot 3^{2\cdot \frac{1}{2}}}\\[5pt] &= 2^{\frac{5}{2}-\frac{1}{2}}\cdot 3^{\frac{1}{2}-1} = 2^{2}\cdot 3^{-\frac{1}{2}} = \frac{4}{\sqrt{3}} = \frac{4\sqrt{3}}{3}\,\textrm{.} \end{align} |
(In the last equality, we multiply top and bottom by \displaystyle \sqrt{3}.)
