Wiki-Quellcode von Lösung Teilweises Wurzelziehen
Version 1.2 von Beate Gomoll am 2025/11/06 13:04
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| author | version | line-number | content |
|---|---|---|---|
| 1 | (%class=123%) | ||
| 2 | 1. | ||
| 3 | ((( | ||
| 4 | (%class=abc%) | ||
| 5 | 1. {{formula}}\sqrt{44}=\sqrt{4\cdot 11}=\sqrt{4}\cdot\sqrt{11}=2 \cdot \sqrt11{{/formula}} | ||
| 6 | 1. {{formula}}\sqrt{75}=\sqrt{25\cdot 3}=\sqrt{25}\cdot\sqrt{3}=5 \cdot \sqrt{3}{{/formula}} | ||
| 7 | 1. {{formula}}\sqrt{63}=\sqrt{9\cdot 7}=\sqrt{9}\cdot\sqrt{7}=3 \cdot \sqrt{7}{{/formula}} | ||
| 8 | 1. {{formula}}\sqrt{98}=\sqrt{49\cdot 2}=\sqrt{49}\cdot\sqrt{2}=7 \cdot \sqrt{2}{{/formula}} | ||
| 9 | ))) | ||
| 10 | 2. | ||
| 11 | ((( | ||
| 12 | (%class=abc%) | ||
| 13 | 1. {{formula}}10 \cdot\sqrt{3}{{/formula}} | ||
| 14 | 1. {{formula}}21 \cdot\sqrt{2}{{/formula}} | ||
| 15 | 1. {{formula}}20 \cdot\sqrt{5}{{/formula}} | ||
| 16 | 1. {{formula}}6 \cdot\sqrt{11}{{/formula}} | ||
| 17 | ))) |