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| lehrkraefte:blc:miniaufgaben:kw03-2017 [2017/01/20 06:51] – created Ivo Blöchliger | lehrkraefte:blc:miniaufgaben:kw03-2017 [2020/08/09 13:16] (current) – external edit 127.0.0.1 | ||
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| + | ==== 16. Januar 2017 bis 21. Januar 2017 ==== | ||
| + | === Dienstag 17. Januar 2017 === | ||
| + | Berechnen Sie: | ||
| + | - $\qquad \left(\frac{4}{3}+2\right)^{-1}\cdot\left(\frac{1}{2}\right)^{-2}$ | ||
| + | - $\qquad \left(-\frac{5}{3}-2\right)^{-1}\cdot\left(\frac{1}{2}\right)^{-1}$ | ||
| + | - $\qquad \left(-\frac{7}{4}-\frac{3}{2}\right)^{-1}\cdot\left(\frac{4}{5}\right)^{-1}$ | ||
| + | | ||
| + | <hidden Lösungen> | ||
| + | - $ \frac{6}{5}$ | ||
| + | - $ -\frac{6}{11}$ | ||
| + | - $ -\frac{5}{13}$ | ||
| + | </ | ||
| + | === Donnerstag 19. Januar 2017 === | ||
| + | Lösen Sie folgende Gleichungen: | ||
| + | - $$\frac{3}{2} \cdot x^{\frac{2}{3}} = 6$$ | ||
| + | - $$\frac{2}{3} \cdot x^{\frac{2}{3}} = 6$$ | ||
| + | - $$\frac{3}{4} \cdot x^{\frac{3}{4}} = 6$$ | ||
| + | <hidden Lösungen> | ||
| + | - $x=8$ | ||
| + | - $x=27$ | ||
| + | - $x=16$ | ||
| + | </ | ||
| + | |||
| + | |||
| + | === Freitag 20. Januar 2017 === | ||
| + | Berechnen Sie von Hand und schreiben Sie in Normalform: | ||
| + | - $\left(\frac{2}{3} + \sqrt{\frac{2}{3}}\right)^2$ | ||
| + | - $\left(\frac{3}{5} + \sqrt{\frac{3}{5}}\right)^2$ | ||
| + | - $\left(\frac{5}{3} + \sqrt{\frac{5}{3}}\right)^2$ | ||
| + | |||
| + | <hidden Lösungen> | ||
| + | - $\frac{4}{9} + 2 \cdot \frac{2}{3}\sqrt{\frac{2}{3}} + \frac{2}{3} = \frac{10}{9} + \frac{4}{9}\sqrt{6}$ | ||
| + | - $\frac{9}{25} + 2 \cdot \frac{3}{5}\sqrt{\frac{3}{5}} + \frac{3}{5} = \frac{24}{25} + \frac{6}{25}\sqrt{15}$ | ||
| + | - $\frac{25}{9} + 2 \cdot \frac{5}{3}\sqrt{\frac{5}{3}} + \frac{5}{3} = \frac{40}{9} + \frac{10}{9}\sqrt{15}$ | ||
| + | </ | ||