miniaufgabe.js
d3.min.js
function-plot.js
==== 24. Oktober 2022 bis 28. Oktober 2022 ====
=== Dienstag 25. Oktober 2022 ===
Ausrechnen, Resultat als gekürzter Bruch:miniAufgabe("#exokettenbruch","#solkettenbruch",
[["$\\displaystyle 2 + \\frac{1}{3 + \\frac{1}{2 + \\frac{1}{4}}}$", "$\\displaystyle 2 + \\frac{1}{3 + \\frac{1}{2 + \\frac{1}{4}}} = 2 + \\frac{1}{3 + \\frac{1}{\\frac{9}{4}}} = 2 + \\frac{1}{3 + \\frac{4}{9}} = 2 + \\frac{1}{\\frac{31}{9}} = 2 + \\frac{9}{31} = \\frac{71}{31}$"], ["$\\displaystyle 2 + \\frac{1}{3 + \\frac{1}{3 + \\frac{1}{3}}}$", "$\\displaystyle 2 + \\frac{1}{3 + \\frac{1}{3 + \\frac{1}{3}}} = 2 + \\frac{1}{3 + \\frac{1}{\\frac{10}{3}}} = 2 + \\frac{1}{3 + \\frac{3}{10}} = 2 + \\frac{1}{\\frac{33}{10}} = 2 + \\frac{10}{33} = \\frac{76}{33}$"], ["$\\displaystyle 1 + \\frac{1}{3 + \\frac{1}{2 + \\frac{1}{3}}}$", "$\\displaystyle 1 + \\frac{1}{3 + \\frac{1}{2 + \\frac{1}{3}}} = 1 + \\frac{1}{3 + \\frac{1}{\\frac{7}{3}}} = 1 + \\frac{1}{3 + \\frac{3}{7}} = 1 + \\frac{1}{\\frac{24}{7}} = 1 + \\frac{7}{24} = \\frac{31}{24}$"], ["$\\displaystyle 2 + \\frac{1}{2 + \\frac{1}{4 + \\frac{1}{4}}}$", "$\\displaystyle 2 + \\frac{1}{2 + \\frac{1}{4 + \\frac{1}{4}}} = 2 + \\frac{1}{2 + \\frac{1}{\\frac{17}{4}}} = 2 + \\frac{1}{2 + \\frac{4}{17}} = 2 + \\frac{1}{\\frac{38}{17}} = 2 + \\frac{17}{38} = \\frac{93}{38}$"], ["$\\displaystyle 1 + \\frac{1}{3 + \\frac{1}{3 + \\frac{1}{4}}}$", "$\\displaystyle 1 + \\frac{1}{3 + \\frac{1}{3 + \\frac{1}{4}}} = 1 + \\frac{1}{3 + \\frac{1}{\\frac{13}{4}}} = 1 + \\frac{1}{3 + \\frac{4}{13}} = 1 + \\frac{1}{\\frac{43}{13}} = 1 + \\frac{13}{43} = \\frac{56}{43}$"], ["$\\displaystyle 1 + \\frac{1}{3 + \\frac{1}{1 + \\frac{1}{5}}}$", "$\\displaystyle 1 + \\frac{1}{3 + \\frac{1}{1 + \\frac{1}{5}}} = 1 + \\frac{1}{3 + \\frac{1}{\\frac{6}{5}}} = 1 + \\frac{1}{3 + \\frac{5}{6}} = 1 + \\frac{1}{\\frac{23}{6}} = 1 + \\frac{6}{23} = \\frac{29}{23}$"], ["$\\displaystyle 1 + \\frac{1}{2 + \\frac{1}{4 + \\frac{1}{3}}}$", "$\\displaystyle 1 + \\frac{1}{2 + \\frac{1}{4 + \\frac{1}{3}}} = 1 + \\frac{1}{2 + \\frac{1}{\\frac{13}{3}}} = 1 + \\frac{1}{2 + \\frac{3}{13}} = 1 + \\frac{1}{\\frac{29}{13}} = 1 + \\frac{13}{29} = \\frac{42}{29}$"], ["$\\displaystyle 2 + \\frac{1}{2 + \\frac{1}{4 + \\frac{1}{3}}}$", "$\\displaystyle 2 + \\frac{1}{2 + \\frac{1}{4 + \\frac{1}{3}}} = 2 + \\frac{1}{2 + \\frac{1}{\\frac{13}{3}}} = 2 + \\frac{1}{2 + \\frac{3}{13}} = 2 + \\frac{1}{\\frac{29}{13}} = 2 + \\frac{13}{29} = \\frac{71}{29}$"], ["$\\displaystyle 1 + \\frac{1}{1 + \\frac{1}{3 + \\frac{1}{5}}}$", "$\\displaystyle 1 + \\frac{1}{1 + \\frac{1}{3 + \\frac{1}{5}}} = 1 + \\frac{1}{1 + \\frac{1}{\\frac{16}{5}}} = 1 + \\frac{1}{1 + \\frac{5}{16}} = 1 + \\frac{1}{\\frac{21}{16}} = 1 + \\frac{16}{21} = \\frac{37}{21}$"], ["$\\displaystyle 1 + \\frac{1}{1 + \\frac{1}{4 + \\frac{1}{5}}}$", "$\\displaystyle 1 + \\frac{1}{1 + \\frac{1}{4 + \\frac{1}{5}}} = 1 + \\frac{1}{1 + \\frac{1}{\\frac{21}{5}}} = 1 + \\frac{1}{1 + \\frac{5}{21}} = 1 + \\frac{1}{\\frac{26}{21}} = 1 + \\frac{21}{26} = \\frac{47}{26}$"]],
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=== Donnerstag 27. Oktober 2022 ===
Lösen Sie nach $x$ auf, ohne Diskussion der Spezialfälle.miniAufgabe("#exoverhaeltnisgleichungen1","#solverhaeltnisgleichungen1",
[["$-\\frac{25}{8}a^{4} : \\frac{25}{14}a^{3} = -\\frac{7}{6}b^{2} : \\frac{2}{5}x$", "$\\begin{align*}\n -\\frac{25}{8}a^{4} : \\frac{25}{14}a^{3} &= -\\frac{7}{6}b^{2} : \\frac{2}{5}x\\\\\n\\frac{2}{5}x \\cdot -\\frac{25}{8}a^{4} &= -\\frac{7}{6}b^{2} \\cdot \\frac{25}{14}a^{3}\\\\\n-\\frac{5}{4}a^{4}x &= -\\frac{25}{12}a^{3}b^{2}\\\\\nx &= \\frac{-\\frac{25}{12}a^{3}b^{2}}{-\\frac{5}{4}a^{4}}\\\\\nx & = \\frac{5}{3}a^{-1}b^{2}\\\\\n\\end{align*}\n$"], ["$-\\frac{5}{8}b^{4} : \\frac{3}{5}b^{3} = -\\frac{25}{2}b^{4} : -\\frac{8}{3}x$", "$\\begin{align*}\n -\\frac{5}{8}b^{4} : \\frac{3}{5}b^{3} &= -\\frac{25}{2}b^{4} : -\\frac{8}{3}x\\\\\n-\\frac{8}{3}x \\cdot -\\frac{5}{8}b^{4} &= \\frac{3}{5}b^{3} \\cdot -\\frac{25}{2}b^{4}\\\\\n\\frac{5}{3}b^{4}x &= -\\frac{15}{2}b^{7}\\\\\nx &= \\frac{-\\frac{15}{2}b^{7}}{\\frac{5}{3}b^{4}}\\\\\nx & = -\\frac{9}{2}b^{3}\\\\\n\\end{align*}\n$"], ["$-\\frac{4}{35}a^{4} : -\\frac{9}{8}b^{2} = -\\frac{16}{35}a^{2} : \\frac{7}{2}x$", "$\\begin{align*}\n -\\frac{4}{35}a^{4} : -\\frac{9}{8}b^{2} &= -\\frac{16}{35}a^{2} : \\frac{7}{2}x\\\\\n\\frac{7}{2}x \\cdot -\\frac{4}{35}a^{4} &= -\\frac{9}{8}b^{2} \\cdot -\\frac{16}{35}a^{2}\\\\\n-\\frac{2}{5}a^{4}x &= \\frac{18}{35}a^{2}b^{2}\\\\\nx &= \\frac{\\frac{18}{35}a^{2}b^{2}}{-\\frac{2}{5}a^{4}}\\\\\nx & = -\\frac{9}{7}a^{-2}b^{2}\\\\\n\\end{align*}\n$"], ["$-\\frac{9}{8}a^{4} : -\\frac{5}{2}x = \\frac{3}{5}b^{4} : \\frac{32}{15}b^{2}$", "$\\begin{align*}\n -\\frac{9}{8}a^{4} : -\\frac{5}{2}x &= \\frac{3}{5}b^{4} : \\frac{32}{15}b^{2}\\\\\n-\\frac{5}{2}x \\cdot \\frac{3}{5}b^{4} &= -\\frac{9}{8}a^{4} \\cdot \\frac{32}{15}b^{2}\\\\\n-\\frac{3}{2}b^{4}x &= -\\frac{12}{5}a^{4}b^{2}\\\\\nx &= \\frac{-\\frac{12}{5}a^{4}b^{2}}{-\\frac{3}{2}b^{4}}\\\\\nx & = \\frac{8}{5}a^{4}b^{-2}\\\\\n\\end{align*}\n$"], ["$-\\frac{1}{2}x : \\frac{28}{27}b^{3} = \\frac{2}{7}b^{3} : -\\frac{8}{3}b^{2}$", "$\\begin{align*}\n -\\frac{1}{2}x : \\frac{28}{27}b^{3} &= \\frac{2}{7}b^{3} : -\\frac{8}{3}b^{2}\\\\\n-\\frac{1}{2}x \\cdot -\\frac{8}{3}b^{2} &= \\frac{2}{7}b^{3} \\cdot \\frac{28}{27}b^{3}\\\\\n\\frac{4}{3}b^{2}x &= \\frac{8}{27}b^{6}\\\\\nx &= \\frac{\\frac{8}{27}b^{6}}{\\frac{4}{3}b^{2}}\\\\\nx & = \\frac{2}{9}b^{4}\\\\\n\\end{align*}\n$"], ["$\\frac{27}{28}a^{2} : -\\frac{4}{5}a^{2} = -\\frac{5}{8}x : -\\frac{2}{3}b^{4}$", "$\\begin{align*}\n \\frac{27}{28}a^{2} : -\\frac{4}{5}a^{2} &= -\\frac{5}{8}x : -\\frac{2}{3}b^{4}\\\\\n-\\frac{5}{8}x \\cdot -\\frac{4}{5}a^{2} &= -\\frac{2}{3}b^{4} \\cdot \\frac{27}{28}a^{2}\\\\\n\\frac{1}{2}a^{2}x &= -\\frac{9}{14}a^{2}b^{4}\\\\\nx &= \\frac{-\\frac{9}{14}a^{2}b^{4}}{\\frac{1}{2}a^{2}}\\\\\nx & = -\\frac{9}{7}b^{4}\\\\\n\\end{align*}\n$"], ["$-\\frac{32}{35}a^{3} : -\\frac{1}{3}b^{2} = \\frac{24}{35}a^{4} : -\\frac{5}{8}x$", "$\\begin{align*}\n -\\frac{32}{35}a^{3} : -\\frac{1}{3}b^{2} &= \\frac{24}{35}a^{4} : -\\frac{5}{8}x\\\\\n-\\frac{5}{8}x \\cdot -\\frac{32}{35}a^{3} &= -\\frac{1}{3}b^{2} \\cdot \\frac{24}{35}a^{4}\\\\\n\\frac{4}{7}a^{3}x &= -\\frac{8}{35}a^{4}b^{2}\\\\\nx &= \\frac{-\\frac{8}{35}a^{4}b^{2}}{\\frac{4}{7}a^{3}}\\\\\nx & = -\\frac{2}{5}ab^{2}\\\\\n\\end{align*}\n$"], ["$-\\frac{7}{5}x : -\\frac{8}{9}a^{2} = -\\frac{27}{32}a^{4} : -\\frac{15}{14}b^{3}$", "$\\begin{align*}\n -\\frac{7}{5}x : -\\frac{8}{9}a^{2} &= -\\frac{27}{32}a^{4} : -\\frac{15}{14}b^{3}\\\\\n-\\frac{7}{5}x \\cdot -\\frac{15}{14}b^{3} &= -\\frac{8}{9}a^{2} \\cdot -\\frac{27}{32}a^{4}\\\\\n\\frac{3}{2}b^{3}x &= \\frac{3}{4}a^{6}\\\\\nx &= \\frac{\\frac{3}{4}a^{6}}{\\frac{3}{2}b^{3}}\\\\\nx & = \\frac{1}{2}a^{6}b^{-3}\\\\\n\\end{align*}\n$"], ["$\\frac{7}{4}b^{2} : \\frac{35}{8}a^{3} = \\frac{2}{5}x : -\\frac{2}{3}a^{4}$", "$\\begin{align*}\n \\frac{7}{4}b^{2} : \\frac{35}{8}a^{3} &= \\frac{2}{5}x : -\\frac{2}{3}a^{4}\\\\\n\\frac{2}{5}x \\cdot \\frac{35}{8}a^{3} &= -\\frac{2}{3}a^{4} \\cdot \\frac{7}{4}b^{2}\\\\\n\\frac{7}{4}a^{3}x &= -\\frac{7}{6}a^{4}b^{2}\\\\\nx &= \\frac{-\\frac{7}{6}a^{4}b^{2}}{\\frac{7}{4}a^{3}}\\\\\nx & = -\\frac{2}{3}ab^{2}\\\\\n\\end{align*}\n$"], ["$\\frac{5}{2}x : -\\frac{9}{32}b^{4} = \\frac{8}{3}a^{4} : \\frac{3}{5}a^{4}$", "$\\begin{align*}\n \\frac{5}{2}x : -\\frac{9}{32}b^{4} &= \\frac{8}{3}a^{4} : \\frac{3}{5}a^{4}\\\\\n\\frac{5}{2}x \\cdot \\frac{3}{5}a^{4} &= \\frac{8}{3}a^{4} \\cdot -\\frac{9}{32}b^{4}\\\\\n\\frac{3}{2}a^{4}x &= -\\frac{3}{4}a^{4}b^{4}\\\\\nx &= \\frac{-\\frac{3}{4}a^{4}b^{4}}{\\frac{3}{2}a^{4}}\\\\\nx & = -\\frac{1}{2}b^{4}\\\\\n\\end{align*}\n$"]],
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ruby verhaeltnis-gleichungen.rb 1