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playground:playground [2018/03/01 10:06] Ivo Blöchliger |
playground:playground [2021/08/02 19:27] (current) Olaf Schnürer old revision restored (2020/07/29 11:38) |
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| ====== PlayGround ====== | ====== PlayGround ====== | ||
| - | $\displaystyle = \left(\mathrm{e}^{x \cdot \left(\frac{19}{12} + \frac{1}{6} - \frac{9}{4}\right)} \right)^{\frac{10}{7}}$$\displaystyle \left(\frac{\mathrm{e}^{-\frac{1}{88}\cdot x} \cdot \left(\mathrm{e}^x\right)^{\frac{12}{11}}}{\sqrt[11]{\mathrm{e}^{-5\cdot x}}}\right)^{-\frac{11}{9}} = \left(\frac{\mathrm{e}^{-\frac{1}{88}\cdot x} \cdot \mathrm{e}^{\frac{12}{11}\cdot x}}{\mathrm{e}^{-\frac{5}{11}\cdot x}}\right)^{-\frac{11}{9}} = \left(\mathrm{e}^{x \cdot \left(-\frac{1}{88} + \frac{12}{11} - \left(-\frac{5}{11}\right)\right)} \right)^{-\frac{11}{9}} = \left(\mathrm{e}^{x \cdot \left(-\frac{1}{88} + \frac{96}{88} - \left(-\frac{40}{88}\right)\right)} \right)^{-\frac{11}{9}} = \left(\mathrm{e}^{x \cdot \frac{135}{88}} \right)^{-\frac{11}{9}} = \left(\mathrm{e}^{x \cdot \frac{135}{88}} \right)^{-\frac{11}{9}} = \mathrm{e}^{x \cdot \frac{135}{88} \cdot \left(-\frac{11}{9}\right)} = \mathrm{e}^{-\frac{15}{8} \cdot x}$ | ||