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--- lehrkraefte:blc:miniaufgaben [2025/12/09 05:49] – Ivo Blöchliger
+++ lehrkraefte:blc:miniaufgaben [2026/02/04 13:00] (current) – [Miniaufgaben] Ivo Blöchliger
@@ Line 3: / Line 3: @@
 ===== Miniaufgaben =====
   * Auf jede Montags-Lektion (ausser Prüfungslektionen) sind eine bis zwei Miniaufgaben vorzubereiten. Am Anfang der Lektion wird ein Würfel geworfen. Zeigt der Würfel eine Vier, Fünf oder Sechs, werden die  Aufgaben in Form eines Kurztests geprüft.
-  * Im ersten Semester gibt <del>drei</del> zwei Joker.
+  * Im zweiten Semester gibt es wieder zwei Joker.
   * Bei Einsatz eines Jokers wird der Schüler vom eventuellen Kurztest ersatzlos dispensiert. Zeigt der Würfel 1-3, ist der Joker aber auch aufgebraucht. Joker werden per e-mail bis 23:59 am Vortag eingelöst werden.
   * Der Minikurztest ist auf mitgebrachtem **A4-Papier im Hochformat** zu lösen. Ausgefranste Ränder, zerknittertes Papier, abgerissene Ecken und Übergrössen führen zu **Abzug**.
@@ Line 16: / Line 16: @@
 </PRELOAD>
+==== Montag 9. Februar 2026 ====
+Ausmultiplizieren und Zusammenfassen.<JS>miniAufgabe("#exoausmult1","#solausmult1",
-==== Montag 15. Dezember 2025 ====
+[["$\\displaystyle \\left(3f^{3}+5k^{2}\\right)\\cdot\\left(-4x^{3}+5k^{2}\\right)-\\left(3f^{3}-5k^{2}\\right)\\cdot\\left(-4x^{3}-5k^{2}\\right)$", "$ \\begin{multline*}-12f^{3}x^{3}+15f^{3}k^{2}-20k^{2}x^{3}+25k^{2}k^{2} - \\left(-12f^{3}x^{3}-15f^{3}k^{2}+20k^{2}x^{3}+25k^{2}k^{2}\\right) = \\\\\n-12f^{3}x^{3}+15f^{3}k^{2}-20k^{2}x^{3}+25k^{2}k^{2} +12f^{3}x^{3}+15f^{3}k^{2}-20k^{2}x^{3}-25k^{2}k^{2} = \\\\\n 30f^{3}k^{2}-40k^{2}x^{3}\\end{multline*}\n$"], ["$\\displaystyle \\left(3w^{3}-4y^{3}\\right)\\cdot\\left(3w^{3}+5e^{2}\\right)-\\left(-3w^{3}-4y^{3}\\right)\\cdot\\left(3w^{3}-5e^{2}\\right)$", "$ \\begin{multline*}9w^{3}w^{3}-12w^{3}y^{3}+15e^{2}w^{3}-20e^{2}y^{3} - \\left(-9w^{3}w^{3}-12w^{3}y^{3}+15e^{2}w^{3}+20e^{2}y^{3}\\right) = \\\\\n9w^{3}w^{3}-12w^{3}y^{3}+15e^{2}w^{3}-20e^{2}y^{3} +9w^{3}w^{3}+12w^{3}y^{3}-15e^{2}w^{3}-20e^{2}y^{3} = \\\\\n 18w^{6}-40e^{2}y^{3}\\end{multline*}\n$"], ["$\\displaystyle \\left(3a^{3}-4b^{3}\\right)\\cdot\\left(3a^{3}+5d^{2}\\right)-\\left(3a^{3}+4b^{3}\\right)\\cdot\\left(-3a^{3}+5d^{2}\\right)$", "$ \\begin{multline*}9a^{3}a^{3}-12a^{3}b^{3}+15a^{3}d^{2}-20b^{3}d^{2} - \\left(-9a^{3}a^{3}-12a^{3}b^{3}+15a^{3}d^{2}+20b^{3}d^{2}\\right) = \\\\\n9a^{3}a^{3}-12a^{3}b^{3}+15a^{3}d^{2}-20b^{3}d^{2} +9a^{3}a^{3}+12a^{3}b^{3}-15a^{3}d^{2}-20b^{3}d^{2} = \\\\\n 18a^{6}-40b^{3}d^{2}\\end{multline*}\n$"], ["$\\displaystyle \\left(4y^{3}-5w^{2}\\right)\\cdot\\left(4y^{3}-3c^{2}\\right)-\\left(-4y^{3}-5w^{2}\\right)\\cdot\\left(-4y^{3}-3c^{2}\\right)$", "$ \\begin{multline*}16y^{3}y^{3}-12c^{2}y^{3}-20w^{2}y^{3}+15c^{2}w^{2} - \\left(16y^{3}y^{3}+12c^{2}y^{3}+20w^{2}y^{3}+15c^{2}w^{2}\\right) = \\\\\n16y^{3}y^{3}-12c^{2}y^{3}-20w^{2}y^{3}+15c^{2}w^{2} -16y^{3}y^{3}-12c^{2}y^{3}-20w^{2}y^{3}-15c^{2}w^{2} = \\\\\n -24c^{2}y^{3}-40w^{2}y^{3}\\end{multline*}\n$"], ["$\\displaystyle \\left(5y^{3}+3c^{2}\\right)\\cdot\\left(-4x^{3}+5y^{3}\\right)-\\left(-5y^{3}+3c^{2}\\right)\\cdot\\left(4x^{3}+5y^{3}\\right)$", "$ \\begin{multline*}-20x^{3}y^{3}+25y^{3}y^{3}-12c^{2}x^{3}+15c^{2}y^{3} - \\left(-20x^{3}y^{3}-25y^{3}y^{3}+12c^{2}x^{3}+15c^{2}y^{3}\\right) = \\\\\n-20x^{3}y^{3}+25y^{3}y^{3}-12c^{2}x^{3}+15c^{2}y^{3} +20x^{3}y^{3}+25y^{3}y^{3}-12c^{2}x^{3}-15c^{2}y^{3} = \\\\\n 50y^{6}-24c^{2}x^{3}\\end{multline*}\n$"], ["$\\displaystyle \\left(5m^{3}+3n^{3}\\right)\\cdot\\left(5m^{3}+4p^{2}\\right)-\\left(5m^{3}-3n^{3}\\right)\\cdot\\left(5m^{3}-4p^{2}\\right)$", "$ \\begin{multline*}25m^{3}m^{3}+15m^{3}n^{3}+20m^{3}p^{2}+12n^{3}p^{2} - \\left(25m^{3}m^{3}-15m^{3}n^{3}-20m^{3}p^{2}+12n^{3}p^{2}\\right) = \\\\\n25m^{3}m^{3}+15m^{3}n^{3}+20m^{3}p^{2}+12n^{3}p^{2} -25m^{3}m^{3}+15m^{3}n^{3}+20m^{3}p^{2}-12n^{3}p^{2} = \\\\\n 30m^{3}n^{3}+40m^{3}p^{2}\\end{multline*}\n$"], ["$\\displaystyle \\left(5h^{3}+3p^{3}\\right)\\cdot\\left(5h^{3}-4b^{2}\\right)-\\left(5h^{3}-3p^{3}\\right)\\cdot\\left(-5h^{3}-4b^{2}\\right)$", "$ \\begin{multline*}25h^{3}h^{3}+15h^{3}p^{3}-20b^{2}h^{3}-12b^{2}p^{3} - \\left(-25h^{3}h^{3}+15h^{3}p^{3}-20b^{2}h^{3}+12b^{2}p^{3}\\right) = \\\\\n25h^{3}h^{3}+15h^{3}p^{3}-20b^{2}h^{3}-12b^{2}p^{3} +25h^{3}h^{3}-15h^{3}p^{3}+20b^{2}h^{3}-12b^{2}p^{3} = \\\\\n 50h^{6}-24b^{2}p^{3}\\end{multline*}\n$"], ["$\\displaystyle \\left(3x^{3}+5n^{2}\\right)\\cdot\\left(-4f^{3}+3x^{3}\\right)-\\left(-3x^{3}+5n^{2}\\right)\\cdot\\left(4f^{3}+3x^{3}\\right)$", "$ \\begin{multline*}-12f^{3}x^{3}+9x^{3}x^{3}-20f^{3}n^{2}+15n^{2}x^{3} - \\left(-12f^{3}x^{3}-9x^{3}x^{3}+20f^{3}n^{2}+15n^{2}x^{3}\\right) = \\\\\n-12f^{3}x^{3}+9x^{3}x^{3}-20f^{3}n^{2}+15n^{2}x^{3} +12f^{3}x^{3}+9x^{3}x^{3}-20f^{3}n^{2}-15n^{2}x^{3} = \\\\\n 18x^{6}-40f^{3}n^{2}\\end{multline*}\n$"], ["$\\displaystyle \\left(-5b^{2}+4c^{2}\\right)\\cdot\\left(-3e^{3}+4c^{2}\\right)-\\left(-5b^{2}-4c^{2}\\right)\\cdot\\left(3e^{3}+4c^{2}\\right)$", "$ \\begin{multline*}15b^{2}e^{3}-12c^{2}e^{3}-20b^{2}c^{2}+16c^{2}c^{2} - \\left(-15b^{2}e^{3}-12c^{2}e^{3}-20b^{2}c^{2}-16c^{2}c^{2}\\right) = \\\\\n15b^{2}e^{3}-12c^{2}e^{3}-20b^{2}c^{2}+16c^{2}c^{2} +15b^{2}e^{3}+12c^{2}e^{3}+20b^{2}c^{2}+16c^{2}c^{2} = \\\\\n 30b^{2}e^{3}+32c^{4}\\end{multline*}\n$"], ["$\\displaystyle \\left(-4b^{3}+5h^{3}\\right)\\cdot\\left(5h^{3}+3a^{2}\\right)-\\left(-4b^{3}-5h^{3}\\right)\\cdot\\left(5h^{3}-3a^{2}\\right)$", "$ \\begin{multline*}-20b^{3}h^{3}+25h^{3}h^{3}-12a^{2}b^{3}+15a^{2}h^{3} - \\left(-20b^{3}h^{3}-25h^{3}h^{3}+12a^{2}b^{3}+15a^{2}h^{3}\\right) = \\\\\n-20b^{3}h^{3}+25h^{3}h^{3}-12a^{2}b^{3}+15a^{2}h^{3} +20b^{3}h^{3}+25h^{3}h^{3}-12a^{2}b^{3}-15a^{2}h^{3} = \\\\\n 50h^{6}-24a^{2}b^{3}\\end{multline*}\n$"]],
-Ausrechnen, Resultat als vollständig gekürzter Bruch. **Achtung:** Potenzen erst ganz am Schluss ausrechnen. Zuerst Basen in Primfaktoren zerlegen und vor dem Multiplizieren kürzen!<JS>miniAufgabe("#exonegativeexponenten","#solnegativeexponenten",
-[["$\\displaystyle \\left(\\frac{4}{3}-\\frac{14}{15}\\right)^{-2} \\cdot \\left(-\\frac{5}{4}\\right)^{-3}$", "$\\displaystyle \\left(\\frac{4}{3}-\\frac{14}{15}\\right)^{-2} \\cdot \\left(-\\frac{5}{4}\\right)^{-3} = \\left(\\frac{20}{15}-\\frac{14}{15}\\right)^{-2} \\cdot \\left(-\\frac{5}{4}\\right)^{-3} = \\left(\\frac{2}{5}\\right)^{-2} \\cdot \\left(-\\frac{4}{5}\\right)^3 = \\left(\\frac{5}{2}\\right)^2 \\cdot \\left(-\\frac{4}{5}\\right)^3 = \\left(\\frac{5}{2}\\right)^2 \\cdot \\left(-\\frac{2^{2}}{5}\\right)^3 = \\frac{5^{2}}{2^{2}} \\cdot -\\frac{2^{6}}{5^{3}} = -\\frac{2^{4}}{5} = -\\frac{16}{5}$"], ["$\\displaystyle \\left(\\frac{1}{2}-\\frac{1}{10}\\right)^{-2} \\cdot \\left(\\frac{5}{6}\\right)^{-3}$", "$\\displaystyle \\left(\\frac{1}{2}-\\frac{1}{10}\\right)^{-2} \\cdot \\left(\\frac{5}{6}\\right)^{-3} = \\left(\\frac{5}{10}-\\frac{1}{10}\\right)^{-2} \\cdot \\left(\\frac{5}{6}\\right)^{-3} = \\left(\\frac{2}{5}\\right)^{-2} \\cdot \\left(\\frac{6}{5}\\right)^3 = \\left(\\frac{5}{2}\\right)^2 \\cdot \\left(\\frac{6}{5}\\right)^3 = \\left(\\frac{5}{2}\\right)^2 \\cdot \\left(\\frac{2 \\cdot 3}{5}\\right)^3 = \\frac{5^{2}}{2^{2}} \\cdot \\frac{2^{3} \\cdot 3^{3}}{5^{3}} = \\frac{2 \\cdot 3^{3}}{5} = \\frac{54}{5}$"], ["$\\displaystyle \\left(-\\frac{6}{7}+\\frac{83}{56}\\right)^{-2} \\cdot \\left(\\frac{6}{5}\\right)^{-3}$", "$\\displaystyle \\left(-\\frac{6}{7}+\\frac{83}{56}\\right)^{-2} \\cdot \\left(\\frac{6}{5}\\right)^{-3} = \\left(-\\frac{48}{56}+\\frac{83}{56}\\right)^{-2} \\cdot \\left(\\frac{6}{5}\\right)^{-3} = \\left(\\frac{5}{8}\\right)^{-2} \\cdot \\left(\\frac{5}{6}\\right)^3 = \\left(\\frac{8}{5}\\right)^2 \\cdot \\left(\\frac{5}{6}\\right)^3 = \\left(\\frac{2^{3}}{5}\\right)^2 \\cdot \\left(\\frac{5}{2 \\cdot 3}\\right)^3 = \\frac{2^{6}}{5^{2}} \\cdot \\frac{5^{3}}{2^{3} \\cdot 3^{3}} = \\frac{2^{3} \\cdot 5}{3^{3}} = \\frac{40}{27}$"], ["$\\displaystyle \\left(\\frac{1}{2}-\\frac{1}{10}\\right)^{-2} \\cdot \\left(-\\frac{5}{6}\\right)^{-3}$", "$\\displaystyle \\left(\\frac{1}{2}-\\frac{1}{10}\\right)^{-2} \\cdot \\left(-\\frac{5}{6}\\right)^{-3} = \\left(\\frac{5}{10}-\\frac{1}{10}\\right)^{-2} \\cdot \\left(-\\frac{5}{6}\\right)^{-3} = \\left(\\frac{2}{5}\\right)^{-2} \\cdot \\left(-\\frac{6}{5}\\right)^3 = \\left(\\frac{5}{2}\\right)^2 \\cdot \\left(-\\frac{6}{5}\\right)^3 = \\left(\\frac{5}{2}\\right)^2 \\cdot \\left(-\\frac{2 \\cdot 3}{5}\\right)^3 = \\frac{5^{2}}{2^{2}} \\cdot -\\frac{2^{3} \\cdot 3^{3}}{5^{3}} = -\\frac{2 \\cdot 3^{3}}{5} = -\\frac{54}{5}$"], ["$\\displaystyle \\left(\\frac{5}{3}-\\frac{34}{15}\\right)^{-2} \\cdot \\left(-\\frac{10}{9}\\right)^{-3}$", "$\\displaystyle \\left(\\frac{5}{3}-\\frac{34}{15}\\right)^{-2} \\cdot \\left(-\\frac{10}{9}\\right)^{-3} = \\left(\\frac{25}{15}-\\frac{34}{15}\\right)^{-2} \\cdot \\left(-\\frac{10}{9}\\right)^{-3} = \\left(-\\frac{3}{5}\\right)^{-2} \\cdot \\left(-\\frac{9}{10}\\right)^3 = \\left(\\frac{5}{3}\\right)^2 \\cdot \\left(-\\frac{9}{10}\\right)^3 = \\left(\\frac{5}{3}\\right)^2 \\cdot \\left(-\\frac{3^{2}}{2 \\cdot 5}\\right)^3 = \\frac{5^{2}}{3^{2}} \\cdot -\\frac{3^{6}}{2^{3} \\cdot 5^{3}} = -\\frac{3^{4}}{2^{3} \\cdot 5} = -\\frac{81}{40}$"], ["$\\displaystyle \\left(-\\frac{6}{5}+\\frac{3}{2}\\right)^{-2} \\cdot \\left(-\\frac{5}{2}\\right)^{-3}$", "$\\displaystyle \\left(-\\frac{6}{5}+\\frac{3}{2}\\right)^{-2} \\cdot \\left(-\\frac{5}{2}\\right)^{-3} = \\left(-\\frac{12}{10}+\\frac{15}{10}\\right)^{-2} \\cdot \\left(-\\frac{5}{2}\\right)^{-3} = \\left(\\frac{3}{10}\\right)^{-2} \\cdot \\left(-\\frac{2}{5}\\right)^3 = \\left(\\frac{10}{3}\\right)^2 \\cdot \\left(-\\frac{2}{5}\\right)^3 = \\left(\\frac{2 \\cdot 5}{3}\\right)^2 \\cdot \\left(-\\frac{2}{5}\\right)^3 = \\frac{2^{2} \\cdot 5^{2}}{3^{2}} \\cdot -\\frac{2^{3}}{5^{3}} = -\\frac{2^{5}}{3^{2} \\cdot 5} = -\\frac{32}{45}$"], ["$\\displaystyle \\left(-\\frac{3}{2}+\\frac{2}{3}\\right)^{-2} \\cdot \\left(-\\frac{3}{5}\\right)^{-3}$", "$\\displaystyle \\left(-\\frac{3}{2}+\\frac{2}{3}\\right)^{-2} \\cdot \\left(-\\frac{3}{5}\\right)^{-3} = \\left(-\\frac{9}{6}+\\frac{4}{6}\\right)^{-2} \\cdot \\left(-\\frac{3}{5}\\right)^{-3} = \\left(-\\frac{5}{6}\\right)^{-2} \\cdot \\left(-\\frac{5}{3}\\right)^3 = \\left(\\frac{6}{5}\\right)^2 \\cdot \\left(-\\frac{5}{3}\\right)^3 = \\left(\\frac{2 \\cdot 3}{5}\\right)^2 \\cdot \\left(-\\frac{5}{3}\\right)^3 = \\frac{2^{2} \\cdot 3^{2}}{5^{2}} \\cdot -\\frac{5^{3}}{3^{3}} = -\\frac{2^{2} \\cdot 5}{3} = -\\frac{20}{3}$"], ["$\\displaystyle \\left(-\\frac{2}{3}+\\frac{3}{2}\\right)^{-2} \\cdot \\left(-\\frac{9}{5}\\right)^{-3}$", "$\\displaystyle \\left(-\\frac{2}{3}+\\frac{3}{2}\\right)^{-2} \\cdot \\left(-\\frac{9}{5}\\right)^{-3} = \\left(-\\frac{4}{6}+\\frac{9}{6}\\right)^{-2} \\cdot \\left(-\\frac{9}{5}\\right)^{-3} = \\left(\\frac{5}{6}\\right)^{-2} \\cdot \\left(-\\frac{5}{9}\\right)^3 = \\left(\\frac{6}{5}\\right)^2 \\cdot \\left(-\\frac{5}{9}\\right)^3 = \\left(\\frac{2 \\cdot 3}{5}\\right)^2 \\cdot \\left(-\\frac{5}{3^{2}}\\right)^3 = \\frac{2^{2} \\cdot 3^{2}}{5^{2}} \\cdot -\\frac{5^{3}}{3^{6}} = -\\frac{2^{2} \\cdot 5}{3^{4}} = -\\frac{20}{81}$"], ["$\\displaystyle \\left(-\\frac{2}{5}+\\frac{4}{35}\\right)^{-2} \\cdot \\left(-\\frac{7}{6}\\right)^{-3}$", "$\\displaystyle \\left(-\\frac{2}{5}+\\frac{4}{35}\\right)^{-2} \\cdot \\left(-\\frac{7}{6}\\right)^{-3} = \\left(-\\frac{14}{35}+\\frac{4}{35}\\right)^{-2} \\cdot \\left(-\\frac{7}{6}\\right)^{-3} = \\left(-\\frac{2}{7}\\right)^{-2} \\cdot \\left(-\\frac{6}{7}\\right)^3 = \\left(\\frac{7}{2}\\right)^2 \\cdot \\left(-\\frac{6}{7}\\right)^3 = \\left(\\frac{7}{2}\\right)^2 \\cdot \\left(-\\frac{2 \\cdot 3}{7}\\right)^3 = \\frac{7^{2}}{2^{2}} \\cdot -\\frac{2^{3} \\cdot 3^{3}}{7^{3}} = -\\frac{2 \\cdot 3^{3}}{7} = -\\frac{54}{7}$"], ["$\\displaystyle \\left(\\frac{3}{4}-\\frac{17}{12}\\right)^{-2} \\cdot \\left(-\\frac{3}{4}\\right)^{-3}$", "$\\displaystyle \\left(\\frac{3}{4}-\\frac{17}{12}\\right)^{-2} \\cdot \\left(-\\frac{3}{4}\\right)^{-3} = \\left(\\frac{9}{12}-\\frac{17}{12}\\right)^{-2} \\cdot \\left(-\\frac{3}{4}\\right)^{-3} = \\left(-\\frac{2}{3}\\right)^{-2} \\cdot \\left(-\\frac{4}{3}\\right)^3 = \\left(\\frac{3}{2}\\right)^2 \\cdot \\left(-\\frac{4}{3}\\right)^3 = \\left(\\frac{3}{2}\\right)^2 \\cdot \\left(-\\frac{2^{2}}{3}\\right)^3 = \\frac{3^{2}}{2^{2}} \\cdot -\\frac{2^{6}}{3^{3}} = -\\frac{2^{4}}{3} = -\\frac{16}{3}$"]],
 " <hr> ", " <hr> ");
 </JS>
 <HTML>
-<div id="exonegativeexponenten"></div>
+<div id="exoausmult1"></div>
 </HTML>
@@ Line 30: / Line 28: @@
 <HTML>
-<div id="solnegativeexponenten"></div>
+<div id="solausmult1"></div>
-<div style='font-size:12px;color:gray;'>ruby potenzen-und-brueche.rb 9</div>
+<div style='font-size:12px;color:gray;'>ruby ausmultiplizieren2.rb 1</div>
 </HTML>
 </hidden>
 ==== Aufgaben vom aktuellen Jahr ====
+  * [[lehrkraefte:blc:miniaufgaben:kw06-2026|KW6, 9. Februar 2026: ]]
   * [[lehrkraefte:blc:miniaufgaben:kw50-2025|KW50, 15. Dezember 2025: Potenzen und Bruchrechnen mit Zahlen]]
   * [[lehrkraefte:blc:miniaufgaben:kw49-2025|KW49, 8. Dezember 2025: Definition Kegelschnitte als geometrische Örter, Potenzen und Bruchrechnen mit Zahlen]]