Lösung 3.1:4c
Aus Online Mathematik Brückenkurs 1
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Each term in the expression can be simplified by breaking down the number under the root sign into its factors, | Each term in the expression can be simplified by breaking down the number under the root sign into its factors, | ||
| - | {{ | + | {{Abgesetzte Formel||<math>\begin{align} |
50 &= 5\cdot 10 = 5\cdot 5\cdot 2 = 2\cdot 5^{2}\,,\\[5pt] | 50 &= 5\cdot 10 = 5\cdot 5\cdot 2 = 2\cdot 5^{2}\,,\\[5pt] | ||
20 &= 2\cdot 10 = 2\cdot 2\cdot 5 = 2^{2}\cdot 5\,,\\[5pt] | 20 &= 2\cdot 10 = 2\cdot 2\cdot 5 = 2^{2}\cdot 5\,,\\[5pt] | ||
| Zeile 10: | Zeile 10: | ||
and then taking the squares out from under the root sign, | and then taking the squares out from under the root sign, | ||
| - | {{ | + | {{Abgesetzte Formel||<math>\begin{align} |
\sqrt{50} &= \sqrt{2\cdot 5^2} = 5\sqrt{2}\,,\\ | \sqrt{50} &= \sqrt{2\cdot 5^2} = 5\sqrt{2}\,,\\ | ||
\sqrt{20} &= \sqrt{2^2\cdot 5} = 2\sqrt{5}\,,\\ | \sqrt{20} &= \sqrt{2^2\cdot 5} = 2\sqrt{5}\,,\\ | ||
| Zeile 19: | Zeile 19: | ||
All together, we get | All together, we get | ||
| - | {{ | + | {{Abgesetzte Formel||<math>\begin{align} |
\sqrt{50} + 4\sqrt{20} - 3\sqrt{18} - 2\sqrt{80} | \sqrt{50} + 4\sqrt{20} - 3\sqrt{18} - 2\sqrt{80} | ||
&= 5\sqrt{2} + 4\cdot 2\sqrt{5} - 3\cdot 3\sqrt{2} - 2\cdot 4\sqrt{5}\\[5pt] | &= 5\sqrt{2} + 4\cdot 2\sqrt{5} - 3\cdot 3\sqrt{2} - 2\cdot 4\sqrt{5}\\[5pt] | ||
Version vom 08:37, 22. Okt. 2008
Each term in the expression can be simplified by breaking down the number under the root sign into its factors,
| \displaystyle \begin{align}
50 &= 5\cdot 10 = 5\cdot 5\cdot 2 = 2\cdot 5^{2}\,,\\[5pt] 20 &= 2\cdot 10 = 2\cdot 2\cdot 5 = 2^{2}\cdot 5\,,\\[5pt] 18 &= 2\cdot 9 = 2\cdot 3\cdot 3 = 2\cdot 3^{2}\,,\\[5pt] 80 &= 8\cdot 10 = (2\cdot 4)\cdot (2\cdot 5) = (2\cdot 2\cdot 2)\cdot (2\cdot 5) = 2^{4}\cdot 5\,, \end{align} |
and then taking the squares out from under the root sign,
| \displaystyle \begin{align}
\sqrt{50} &= \sqrt{2\cdot 5^2} = 5\sqrt{2}\,,\\ \sqrt{20} &= \sqrt{2^2\cdot 5} = 2\sqrt{5}\,,\\ \sqrt{18} &= \sqrt{2\cdot 3^2} = 3\sqrt{2}\,,\\ \sqrt{80} &= \sqrt{2^4\cdot 5} = 2^{2}\sqrt{5} = 4\sqrt{5}\,\textrm{.} \end{align} |
All together, we get
| \displaystyle \begin{align}
\sqrt{50} + 4\sqrt{20} - 3\sqrt{18} - 2\sqrt{80} &= 5\sqrt{2} + 4\cdot 2\sqrt{5} - 3\cdot 3\sqrt{2} - 2\cdot 4\sqrt{5}\\[5pt] &= 5\sqrt{2} + 8\sqrt{5} - 9\sqrt{2} - 8\sqrt{5}\\[5pt] &= (5-9)\sqrt{2} + (8-8)\sqrt{5} = -4\sqrt{2}\,\textrm{.} \end{align} |
