lehrkraefte:blc:miniaufgaben:kw47-2023 [https://fginfo.ksbg.ch Fachgruppe Informatik der KSBG]

This page is read only. You can view the source, but not change it. Ask your administrator if you think this is wrong.
<PRELOAD>
  miniaufgabe.js
</PRELOAD>

==== 20. November 2023 bis 24. November 2023 ====
=== Montag 20. November 2023 ===
Schreiben Sie das entsprechende Potenzgesetz auf und beweisen Sie es für natürliche Zahlen:<JS>miniAufgabe("#exopotenzgesetzebeweisen","#solpotenzgesetzebeweisen",
[["$a^n \\cdot a^m$", "$a^n\\cdot a^m = a^{n+m}$. \n     <b>Beweis</b>: $a^n \\cdot a^m = \n    \\underbrace{a\\cdot a \\cdot \\ldots \\cdot a}_{\\text{Anzahl Faktoren: } n} \\cdot\n    \\underbrace{a\\cdot a \\cdot \\ldots \\cdot a}_{\\text{Anzahl Faktoren: } m } =\n    \\underbrace{a\\cdot a \\cdot \\ldots \\cdot a}_{\\text{Anzahl Faktoren: } n+m} =a^{n+m}$"], ["$(a \\cdot b)^n$", "$(a \\cdot b)^n = a^n \\cdot b^n$. <b>Beweis</b>: $(a \\cdot b)^n = \n    \\underbrace{(a \\cdot b) \\cdot (a \\cdot b) \\cdot \\ldots \\cdot (a \\cdot b)}_{\\text{Anzahl Klammern: } n } =\n    \\underbrace{a\\cdot a \\cdot \\ldots \\cdot a}_{\\text{Anzahl Faktoren: } n } \\cdot\n    \\underbrace{b\\cdot b \\cdot \\ldots \\cdot b}_{\\text{Anzahl Faktoren: } n} = a^n \\cdot b^n$."], ["$\\left(a^n\\right)^m$", "$\\left(a^n\\right)^m = a^{n \\cdot m}$.\n    <b>Beweis</b>: $\\left(a^n\\right)^m = \n    \\underbrace{a^n\\cdot a^n \\cdot \\ldots \\cdot a^n}_{\\text{Anzahl Potenzen: } m } =\n    \\underbrace{\\left(\\underbrace{a\\cdot a \\cdot \\ldots \\cdot a}_{\\text{Anzahl Faktoren: } n}\\right)\\cdot \n        \\left(\\underbrace{a\\cdot a \\cdot \\ldots \\cdot a}_{\\text{Anzahl Faktoren: } n}\\right) \\cdot \\ldots \\cdot\n        \\left(\\underbrace{a\\cdot a \\cdot \\ldots \\cdot a}_{\\text{Anzahl Faktoren: } n}\\right)}_{\\text{Anzahl Klammern: } m} =\n    a^{n \\cdot m}$."], ["$\\frac{a^n}{a^m}$ für $n \\geq m$", "$\\frac{a^n}{a^m} = a^{n-m}$. <b>Beweis</b>\n    $\\frac{a^n}{a^m} = \\frac{ \\overbrace{a\\cdot a \\cdot \\ldots \\cdot a}^{\\text{Anzahl Faktoren: } n}}{\\underbrace{a\\cdot a \\cdot \\ldots \\cdot a}_{\\text{Anzahl Faktoren: } m }} \\stackrel{m \\text{Faktoren kürzen}}{=} \\frac{\\overbrace{a\\cdot a \\cdot \\ldots \\cdot a}^{\\text{Anzahl Faktoren: } n-m }}{1} = a^{n-m}$."], ["$\\left(\\frac{a}{b}\\right)^n$", "$\\left(\\frac{a}{b}\\right)^n = \\frac{a^n}{b^n}$. <b>Beweis</b>:\n    $\\left(\\frac{a}{b}\\right)^n = \\underbrace{\\frac{a}{b}\\cdot \\frac{a}{b} \\cdot \\ldots \\cdot \\frac{a}{b}}_{\\text{Anzahl Brüche: } n } =\n    \\frac{\\overbrace{a\\cdot a \\cdot \\ldots \\cdot a}^{\\text{Anzahl Faktoren: } n}}{\\underbrace{b\\cdot b \\cdot \\ldots \\cdot b}_{\\text{Anzahl Faktoren: } n }} =\n    \\frac{a^n}{b^n}$."]],
" <br><hr> ", "<br><hr>");
</JS>
<HTML>
<div id="exopotenzgesetzebeweisen"></div>

</HTML>
<hidden Lösungen>

<HTML>
<div id="solpotenzgesetzebeweisen"></div>
<div style='font-size:12px;color:gray;'>ruby potenzgesetzebeweisen.rb </div>
</HTML>

</hidden>


=== Dienstag 21. November 2023 ===
Resultat als gekürzten Bruch von Produkten von Potenzen mit positiven Exponenten.<JS>miniAufgabe("#exopotenzenBruecheVariablen1","#solpotenzenBruecheVariablen1",
[["$\\displaystyle \\left(\\frac{\\left(p^{-5} \\cdot d^{2}\\right)^{-3}}{\\left(d^{2} \\cdot x^{-4}\\right)^{4}}\\right)^{-4}$", "$\\left(\\frac{\\left(p^{-5} \\cdot d^{2}\\right)^{-3}}{\\left(d^{2} \\cdot x^{-4}\\right)^{4}}\\right)^{-4} = \\left(\\frac{\\left(p^{-5}\\right)^{-3} \\cdot \\left(d^{2}\\right)^{-3}}{\\left(d^{2}\\right)^{4} \\cdot \\left(x^{-4}\\right)^{4}}\\right)^{-4} = \\left(\\frac{p^{-5\\left(-3\\right)} \\cdot d^{2\\left(-3\\right)}}{d^{2 \\cdot 4} \\cdot x^{-4 \\cdot 4}}\\right)^{-4} = \\left(\\frac{p^{15} \\cdot d^{-6}}{d^{8} \\cdot x^{-16}}\\right)^{-4} = \\left(\\frac{p^{15}}{d^{14}x^{-16}}\\right)^{-4} = \\frac{\\left(p^{15}\\right)^{-4}}{\\left(d^{14}x^{-16}\\right)^{-4}} = \\frac{p^{15\\left(-4\\right)}}{\\left(d^{14}\\right)^{-4} \\cdot \\left(x^{-16}\\right)^{-4}} = \\frac{p^{-60}}{d^{14\\left(-4\\right)} \\cdot x^{-16\\left(-4\\right)}} = \\frac{p^{-60}}{d^{-56} \\cdot x^{64}} = p^{-60}d^{56}x^{-64} = d^{56}p^{-60}x^{-64} = \\frac{d^{56}}{p^{60}x^{64}}$"], ["$\\displaystyle \\left(\\frac{\\left(b^{-4} \\cdot k^{-4}\\right)^{4}}{\\left(k^{5} \\cdot n^{-4}\\right)^{-2}}\\right)^{4}$", "$\\left(\\frac{\\left(b^{-4} \\cdot k^{-4}\\right)^{4}}{\\left(k^{5} \\cdot n^{-4}\\right)^{-2}}\\right)^{4} = \\left(\\frac{\\left(b^{-4}\\right)^{4} \\cdot \\left(k^{-4}\\right)^{4}}{\\left(k^{5}\\right)^{-2} \\cdot \\left(n^{-4}\\right)^{-2}}\\right)^{4} = \\left(\\frac{b^{-4 \\cdot 4} \\cdot k^{-4 \\cdot 4}}{k^{5\\left(-2\\right)} \\cdot n^{-4\\left(-2\\right)}}\\right)^{4} = \\left(\\frac{b^{-16} \\cdot k^{-16}}{k^{-10} \\cdot n^{8}}\\right)^{4} = \\left(\\frac{b^{-16}}{k^{6}n^{8}}\\right)^{4} = \\frac{\\left(b^{-16}\\right)^{4}}{\\left(k^{6}n^{8}\\right)^{4}} = \\frac{b^{-16 \\cdot 4}}{\\left(k^{6}\\right)^{4} \\cdot \\left(n^{8}\\right)^{4}} = \\frac{b^{-64}}{k^{6 \\cdot 4} \\cdot n^{8 \\cdot 4}} = \\frac{b^{-64}}{k^{24} \\cdot n^{32}} = b^{-64}k^{-24}n^{-32} = \\frac{1}{b^{64}k^{24}n^{32}}$"], ["$\\displaystyle \\left(\\frac{\\left(k^{-3} \\cdot e^{3}\\right)^{-3}}{\\left(e^{-5} \\cdot h^{-3}\\right)^{-5}}\\right)^{2}$", "$\\left(\\frac{\\left(k^{-3} \\cdot e^{3}\\right)^{-3}}{\\left(e^{-5} \\cdot h^{-3}\\right)^{-5}}\\right)^{2} = \\left(\\frac{\\left(k^{-3}\\right)^{-3} \\cdot \\left(e^{3}\\right)^{-3}}{\\left(e^{-5}\\right)^{-5} \\cdot \\left(h^{-3}\\right)^{-5}}\\right)^{2} = \\left(\\frac{k^{-3\\left(-3\\right)} \\cdot e^{3\\left(-3\\right)}}{e^{-5\\left(-5\\right)} \\cdot h^{-3\\left(-5\\right)}}\\right)^{2} = \\left(\\frac{k^{9} \\cdot e^{-9}}{e^{25} \\cdot h^{15}}\\right)^{2} = \\left(\\frac{k^{9}}{e^{34}h^{15}}\\right)^{2} = \\frac{\\left(k^{9}\\right)^{2}}{\\left(e^{34}h^{15}\\right)^{2}} = \\frac{k^{9 \\cdot 2}}{\\left(e^{34}\\right)^{2} \\cdot \\left(h^{15}\\right)^{2}} = \\frac{k^{18}}{e^{34 \\cdot 2} \\cdot h^{15 \\cdot 2}} = \\frac{k^{18}}{e^{68} \\cdot h^{30}} = k^{18}e^{-68}h^{-30} = e^{-68}h^{-30}k^{18} = \\frac{k^{18}}{e^{68}h^{30}}$"], ["$\\displaystyle \\left(\\frac{\\left(x^{4} \\cdot h^{2}\\right)^{-3}}{\\left(h^{4} \\cdot b^{-3}\\right)^{4}}\\right)^{2}$", "$\\left(\\frac{\\left(x^{4} \\cdot h^{2}\\right)^{-3}}{\\left(h^{4} \\cdot b^{-3}\\right)^{4}}\\right)^{2} = \\left(\\frac{\\left(x^{4}\\right)^{-3} \\cdot \\left(h^{2}\\right)^{-3}}{\\left(h^{4}\\right)^{4} \\cdot \\left(b^{-3}\\right)^{4}}\\right)^{2} = \\left(\\frac{x^{4\\left(-3\\right)} \\cdot h^{2\\left(-3\\right)}}{h^{4 \\cdot 4} \\cdot b^{-3 \\cdot 4}}\\right)^{2} = \\left(\\frac{x^{-12} \\cdot h^{-6}}{h^{16} \\cdot b^{-12}}\\right)^{2} = \\left(\\frac{x^{-12}}{b^{-12}h^{22}}\\right)^{2} = \\frac{\\left(x^{-12}\\right)^{2}}{\\left(b^{-12}h^{22}\\right)^{2}} = \\frac{x^{-12 \\cdot 2}}{\\left(b^{-12}\\right)^{2} \\cdot \\left(h^{22}\\right)^{2}} = \\frac{x^{-24}}{b^{-12 \\cdot 2} \\cdot h^{22 \\cdot 2}} = \\frac{x^{-24}}{b^{-24} \\cdot h^{44}} = x^{-24}b^{24}h^{-44} = b^{24}h^{-44}x^{-24} = \\frac{b^{24}}{h^{44}x^{24}}$"], ["$\\displaystyle \\left(\\frac{\\left(c^{5} \\cdot e^{-3}\\right)^{4}}{\\left(e^{-3} \\cdot x^{2}\\right)^{3}}\\right)^{3}$", "$\\left(\\frac{\\left(c^{5} \\cdot e^{-3}\\right)^{4}}{\\left(e^{-3} \\cdot x^{2}\\right)^{3}}\\right)^{3} = \\left(\\frac{\\left(c^{5}\\right)^{4} \\cdot \\left(e^{-3}\\right)^{4}}{\\left(e^{-3}\\right)^{3} \\cdot \\left(x^{2}\\right)^{3}}\\right)^{3} = \\left(\\frac{c^{5 \\cdot 4} \\cdot e^{-3 \\cdot 4}}{e^{-3 \\cdot 3} \\cdot x^{2 \\cdot 3}}\\right)^{3} = \\left(\\frac{c^{20} \\cdot e^{-12}}{e^{-9} \\cdot x^{6}}\\right)^{3} = \\left(\\frac{c^{20}}{e^{3}x^{6}}\\right)^{3} = \\frac{\\left(c^{20}\\right)^{3}}{\\left(e^{3}x^{6}\\right)^{3}} = \\frac{c^{20 \\cdot 3}}{\\left(e^{3}\\right)^{3} \\cdot \\left(x^{6}\\right)^{3}} = \\frac{c^{60}}{e^{3 \\cdot 3} \\cdot x^{6 \\cdot 3}} = \\frac{c^{60}}{e^{9} \\cdot x^{18}} = c^{60}e^{-9}x^{-18} = \\frac{c^{60}}{e^{9}x^{18}}$"], ["$\\displaystyle \\left(\\frac{\\left(n^{3} \\cdot k^{-4}\\right)^{5}}{\\left(k^{2} \\cdot y^{4}\\right)^{-5}}\\right)^{5}$", "$\\left(\\frac{\\left(n^{3} \\cdot k^{-4}\\right)^{5}}{\\left(k^{2} \\cdot y^{4}\\right)^{-5}}\\right)^{5} = \\left(\\frac{\\left(n^{3}\\right)^{5} \\cdot \\left(k^{-4}\\right)^{5}}{\\left(k^{2}\\right)^{-5} \\cdot \\left(y^{4}\\right)^{-5}}\\right)^{5} = \\left(\\frac{n^{3 \\cdot 5} \\cdot k^{-4 \\cdot 5}}{k^{2\\left(-5\\right)} \\cdot y^{4\\left(-5\\right)}}\\right)^{5} = \\left(\\frac{n^{15} \\cdot k^{-20}}{k^{-10} \\cdot y^{-20}}\\right)^{5} = \\left(\\frac{n^{15}}{k^{10}y^{-20}}\\right)^{5} = \\frac{\\left(n^{15}\\right)^{5}}{\\left(k^{10}y^{-20}\\right)^{5}} = \\frac{n^{15 \\cdot 5}}{\\left(k^{10}\\right)^{5} \\cdot \\left(y^{-20}\\right)^{5}} = \\frac{n^{75}}{k^{10 \\cdot 5} \\cdot y^{-20 \\cdot 5}} = \\frac{n^{75}}{k^{50} \\cdot y^{-100}} = n^{75}k^{-50}y^{100} = k^{-50}n^{75}y^{100} = \\frac{n^{75}y^{100}}{k^{50}}$"], ["$\\displaystyle \\left(\\frac{\\left(c^{2} \\cdot b^{2}\\right)^{4}}{\\left(b^{-2} \\cdot y^{-2}\\right)^{3}}\\right)^{-4}$", "$\\left(\\frac{\\left(c^{2} \\cdot b^{2}\\right)^{4}}{\\left(b^{-2} \\cdot y^{-2}\\right)^{3}}\\right)^{-4} = \\left(\\frac{\\left(c^{2}\\right)^{4} \\cdot \\left(b^{2}\\right)^{4}}{\\left(b^{-2}\\right)^{3} \\cdot \\left(y^{-2}\\right)^{3}}\\right)^{-4} = \\left(\\frac{c^{2 \\cdot 4} \\cdot b^{2 \\cdot 4}}{b^{-2 \\cdot 3} \\cdot y^{-2 \\cdot 3}}\\right)^{-4} = \\left(\\frac{c^{8} \\cdot b^{8}}{b^{-6} \\cdot y^{-6}}\\right)^{-4} = \\left(\\frac{b^{14}c^{8}}{y^{-6}}\\right)^{-4} = \\frac{\\left(b^{14}c^{8}\\right)^{-4}}{\\left(y^{-6}\\right)^{-4}} = \\frac{\\left(b^{14}\\right)^{-4} \\cdot \\left(c^{8}\\right)^{-4}}{y^{-6\\left(-4\\right)}} = \\frac{b^{14\\left(-4\\right)} \\cdot c^{8\\left(-4\\right)}}{y^{24}} = \\frac{b^{-56} \\cdot c^{-32}}{y^{24}} = b^{-56}c^{-32}y^{-24} = \\frac{1}{b^{56}c^{32}y^{24}}$"], ["$\\displaystyle \\left(\\frac{\\left(a^{3} \\cdot e^{-3}\\right)^{-3}}{\\left(e^{-2} \\cdot x^{4}\\right)^{2}}\\right)^{-3}$", "$\\left(\\frac{\\left(a^{3} \\cdot e^{-3}\\right)^{-3}}{\\left(e^{-2} \\cdot x^{4}\\right)^{2}}\\right)^{-3} = \\left(\\frac{\\left(a^{3}\\right)^{-3} \\cdot \\left(e^{-3}\\right)^{-3}}{\\left(e^{-2}\\right)^{2} \\cdot \\left(x^{4}\\right)^{2}}\\right)^{-3} = \\left(\\frac{a^{3\\left(-3\\right)} \\cdot e^{-3\\left(-3\\right)}}{e^{-2 \\cdot 2} \\cdot x^{4 \\cdot 2}}\\right)^{-3} = \\left(\\frac{a^{-9} \\cdot e^{9}}{e^{-4} \\cdot x^{8}}\\right)^{-3} = \\left(\\frac{a^{-9}e^{13}}{x^{8}}\\right)^{-3} = \\frac{\\left(a^{-9}e^{13}\\right)^{-3}}{\\left(x^{8}\\right)^{-3}} = \\frac{\\left(a^{-9}\\right)^{-3} \\cdot \\left(e^{13}\\right)^{-3}}{x^{8\\left(-3\\right)}} = \\frac{a^{-9\\left(-3\\right)} \\cdot e^{13\\left(-3\\right)}}{x^{-24}} = \\frac{a^{27} \\cdot e^{-39}}{x^{-24}} = a^{27}e^{-39}x^{24} = \\frac{a^{27}x^{24}}{e^{39}}$"], ["$\\displaystyle \\left(\\frac{\\left(n^{2} \\cdot e^{3}\\right)^{4}}{\\left(e^{5} \\cdot b^{3}\\right)^{2}}\\right)^{-5}$", "$\\left(\\frac{\\left(n^{2} \\cdot e^{3}\\right)^{4}}{\\left(e^{5} \\cdot b^{3}\\right)^{2}}\\right)^{-5} = \\left(\\frac{\\left(n^{2}\\right)^{4} \\cdot \\left(e^{3}\\right)^{4}}{\\left(e^{5}\\right)^{2} \\cdot \\left(b^{3}\\right)^{2}}\\right)^{-5} = \\left(\\frac{n^{2 \\cdot 4} \\cdot e^{3 \\cdot 4}}{e^{5 \\cdot 2} \\cdot b^{3 \\cdot 2}}\\right)^{-5} = \\left(\\frac{n^{8} \\cdot e^{12}}{e^{10} \\cdot b^{6}}\\right)^{-5} = \\left(\\frac{e^{2}n^{8}}{b^{6}}\\right)^{-5} = \\frac{\\left(e^{2}n^{8}\\right)^{-5}}{\\left(b^{6}\\right)^{-5}} = \\frac{\\left(e^{2}\\right)^{-5} \\cdot \\left(n^{8}\\right)^{-5}}{b^{6\\left(-5\\right)}} = \\frac{e^{2\\left(-5\\right)} \\cdot n^{8\\left(-5\\right)}}{b^{-30}} = \\frac{e^{-10} \\cdot n^{-40}}{b^{-30}} = e^{-10}n^{-40}b^{30} = b^{30}e^{-10}n^{-40} = \\frac{b^{30}}{e^{10}n^{40}}$"], ["$\\displaystyle \\left(\\frac{\\left(d^{4} \\cdot m^{3}\\right)^{2}}{\\left(m^{-3} \\cdot p^{2}\\right)^{4}}\\right)^{3}$", "$\\left(\\frac{\\left(d^{4} \\cdot m^{3}\\right)^{2}}{\\left(m^{-3} \\cdot p^{2}\\right)^{4}}\\right)^{3} = \\left(\\frac{\\left(d^{4}\\right)^{2} \\cdot \\left(m^{3}\\right)^{2}}{\\left(m^{-3}\\right)^{4} \\cdot \\left(p^{2}\\right)^{4}}\\right)^{3} = \\left(\\frac{d^{4 \\cdot 2} \\cdot m^{3 \\cdot 2}}{m^{-3 \\cdot 4} \\cdot p^{2 \\cdot 4}}\\right)^{3} = \\left(\\frac{d^{8} \\cdot m^{6}}{m^{-12} \\cdot p^{8}}\\right)^{3} = \\left(\\frac{d^{8}m^{18}}{p^{8}}\\right)^{3} = \\frac{\\left(d^{8}m^{18}\\right)^{3}}{\\left(p^{8}\\right)^{3}} = \\frac{\\left(d^{8}\\right)^{3} \\cdot \\left(m^{18}\\right)^{3}}{p^{8 \\cdot 3}} = \\frac{d^{8 \\cdot 3} \\cdot m^{18 \\cdot 3}}{p^{24}} = \\frac{d^{24} \\cdot m^{54}}{p^{24}} = d^{24}m^{54}p^{-24} = \\frac{d^{24}m^{54}}{p^{24}}$"]],
" <hr> ", " <hr> ");
</JS>
<HTML>
<div id="exopotenzenBruecheVariablen1"></div>

</HTML>
<hidden Lösungen>

<HTML>
<div id="solpotenzenBruecheVariablen1"></div>
<div style='font-size:12px;color:gray;'>ruby potenzen-und-brueche-variablen.rb 1</div>
</HTML>

</hidden>