miniaufgabe.js ==== 20. Januar 2020 bis 24. Januar 2020 ==== === Dienstag 21. Januar 2020 === Auf einen Bruchstrich, zusammenfassen, faktorisieren, kürzen.miniAufgabe("#exopoly_bruch_add_faktor","#solpoly_bruch_add_faktor", [["$\\displaystyle \\frac{9x^2+14x+8}{3x^2-12} + \\frac{4}{3x+6}$", "$\\begin{multline*} \\frac{9x^2+14x+8}{3x^2-12} + \\frac{4}{3x+6} = \\frac{9x^2+14x+8}{3(x^2-4)} + \\frac{4}{3(x+2)} = \\frac{9x^2+14x+8}{3(x+2)(x-2)} + \\frac{4(x-2)}{3(x+2)(x-2)} = \\\\ \\frac{9x^2+14x+8}{3(x^2-4)} + \\frac{4x-8}{3(x^2-4)} = \\frac{9x^2+18x}{3(x^2-4)} = \\frac{9x(x+2)}{3(x^2-4)} = \\\\ \\frac{3\\cdot3x(x+2)}{3(x^2-4)} = \\frac{3x(x+2)}{(x+2)(x-2)} = \\frac{3x}{x-2} \\end{multline*}$"], ["$\\displaystyle \\frac{20x^2-46x-12}{5x^2-20} + \\frac{6}{5x-10}$", "$\\begin{multline*} \\frac{20x^2-46x-12}{5x^2-20} + \\frac{6}{5x-10} = \\frac{20x^2-46x-12}{5(x^2-4)} + \\frac{6}{5(x-2)} = \\frac{20x^2-46x-12}{5(x+2)(x-2)} + \\frac{6(x+2)}{5(x+2)(x-2)} = \\\\ \\frac{20x^2-46x-12}{5(x^2-4)} + \\frac{6x+12}{5(x^2-4)} = \\frac{20x^2-40x}{5(x^2-4)} = \\frac{20x(x-2)}{5(x^2-4)} = \\\\ \\frac{5\\cdot4x(x-2)}{5(x^2-4)} = \\frac{4x(x-2)}{(x+2)(x-2)} = \\frac{4x}{x+2} \\end{multline*}$"], ["$\\displaystyle \\frac{8x^2+21x+9}{2x^2-18} + \\frac{3}{2x+6}$", "$\\begin{multline*} \\frac{8x^2+21x+9}{2x^2-18} + \\frac{3}{2x+6} = \\frac{8x^2+21x+9}{2(x^2-9)} + \\frac{3}{2(x+3)} = \\frac{8x^2+21x+9}{2(x+3)(x-3)} + \\frac{3(x-3)}{2(x+3)(x-3)} = \\\\ \\frac{8x^2+21x+9}{2(x^2-9)} + \\frac{3x-9}{2(x^2-9)} = \\frac{8x^2+24x}{2(x^2-9)} = \\frac{8x(x+3)}{2(x^2-9)} = \\\\ \\frac{2\\cdot4x(x+3)}{2(x^2-9)} = \\frac{4x(x+3)}{(x+3)(x-3)} = \\frac{4x}{x-3} \\end{multline*}$"], ["$\\displaystyle \\frac{9x^2-50x-25}{3x^2-75} + \\frac{5}{3x-15}$", "$\\begin{multline*} \\frac{9x^2-50x-25}{3x^2-75} + \\frac{5}{3x-15} = \\frac{9x^2-50x-25}{3(x^2-25)} + \\frac{5}{3(x-5)} = \\frac{9x^2-50x-25}{3(x+5)(x-5)} + \\frac{5(x+5)}{3(x+5)(x-5)} = \\\\ \\frac{9x^2-50x-25}{3(x^2-25)} + \\frac{5x+25}{3(x^2-25)} = \\frac{9x^2-45x}{3(x^2-25)} = \\frac{9x(x-5)}{3(x^2-25)} = \\\\ \\frac{3\\cdot3x(x-5)}{3(x^2-25)} = \\frac{3x(x-5)}{(x+5)(x-5)} = \\frac{3x}{x+5} \\end{multline*}$"], ["$\\displaystyle \\frac{20x^2+77x+12}{4x^2-64} + \\frac{3}{4x+16}$", "$\\begin{multline*} \\frac{20x^2+77x+12}{4x^2-64} + \\frac{3}{4x+16} = \\frac{20x^2+77x+12}{4(x^2-16)} + \\frac{3}{4(x+4)} = \\frac{20x^2+77x+12}{4(x+4)(x-4)} + \\frac{3(x-4)}{4(x+4)(x-4)} = \\\\ \\frac{20x^2+77x+12}{4(x^2-16)} + \\frac{3x-12}{4(x^2-16)} = \\frac{20x^2+80x}{4(x^2-16)} = \\frac{20x(x+4)}{4(x^2-16)} = \\\\ \\frac{4\\cdot5x(x+4)}{4(x^2-16)} = \\frac{5x(x+4)}{(x+4)(x-4)} = \\frac{5x}{x-4} \\end{multline*}$"], ["$\\displaystyle \\frac{10x^2-35x-15}{2x^2-18} + \\frac{5}{2x-6}$", "$\\begin{multline*} \\frac{10x^2-35x-15}{2x^2-18} + \\frac{5}{2x-6} = \\frac{10x^2-35x-15}{2(x^2-9)} + \\frac{5}{2(x-3)} = \\frac{10x^2-35x-15}{2(x+3)(x-3)} + \\frac{5(x+3)}{2(x+3)(x-3)} = \\\\ \\frac{10x^2-35x-15}{2(x^2-9)} + \\frac{5x+15}{2(x^2-9)} = \\frac{10x^2-30x}{2(x^2-9)} = \\frac{10x(x-3)}{2(x^2-9)} = \\\\ \\frac{2\\cdot5x(x-3)}{2(x^2-9)} = \\frac{5x(x-3)}{(x+3)(x-3)} = \\frac{5x}{x+3} \\end{multline*}$"], ["$\\displaystyle \\frac{15x^2+28x+4}{5x^2-20} + \\frac{2}{5x+10}$", "$\\begin{multline*} \\frac{15x^2+28x+4}{5x^2-20} + \\frac{2}{5x+10} = \\frac{15x^2+28x+4}{5(x^2-4)} + \\frac{2}{5(x+2)} = \\frac{15x^2+28x+4}{5(x+2)(x-2)} + \\frac{2(x-2)}{5(x+2)(x-2)} = \\\\ \\frac{15x^2+28x+4}{5(x^2-4)} + \\frac{2x-4}{5(x^2-4)} = \\frac{15x^2+30x}{5(x^2-4)} = \\frac{15x(x+2)}{5(x^2-4)} = \\\\ \\frac{5\\cdot3x(x+2)}{5(x^2-4)} = \\frac{3x(x+2)}{(x+2)(x-2)} = \\frac{3x}{x-2} \\end{multline*}$"], ["$\\displaystyle \\frac{4x^2-15x-9}{2x^2-18} + \\frac{3}{2x-6}$", "$\\begin{multline*} \\frac{4x^2-15x-9}{2x^2-18} + \\frac{3}{2x-6} = \\frac{4x^2-15x-9}{2(x^2-9)} + \\frac{3}{2(x-3)} = \\frac{4x^2-15x-9}{2(x+3)(x-3)} + \\frac{3(x+3)}{2(x+3)(x-3)} = \\\\ \\frac{4x^2-15x-9}{2(x^2-9)} + \\frac{3x+9}{2(x^2-9)} = \\frac{4x^2-12x}{2(x^2-9)} = \\frac{4x(x-3)}{2(x^2-9)} = \\\\ \\frac{2\\cdot2x(x-3)}{2(x^2-9)} = \\frac{2x(x-3)}{(x+3)(x-3)} = \\frac{2x}{x+3} \\end{multline*}$"], ["$\\displaystyle \\frac{9x^2-20x-4}{3x^2-12} + \\frac{2}{3x-6}$", "$\\begin{multline*} \\frac{9x^2-20x-4}{3x^2-12} + \\frac{2}{3x-6} = \\frac{9x^2-20x-4}{3(x^2-4)} + \\frac{2}{3(x-2)} = \\frac{9x^2-20x-4}{3(x+2)(x-2)} + \\frac{2(x+2)}{3(x+2)(x-2)} = \\\\ \\frac{9x^2-20x-4}{3(x^2-4)} + \\frac{2x+4}{3(x^2-4)} = \\frac{9x^2-18x}{3(x^2-4)} = \\frac{9x(x-2)}{3(x^2-4)} = \\\\ \\frac{3\\cdot3x(x-2)}{3(x^2-4)} = \\frac{3x(x-2)}{(x+2)(x-2)} = \\frac{3x}{x+2} \\end{multline*}$"], ["$\\displaystyle \\frac{9x^2+25x+6}{3x^2-27} + \\frac{2}{3x+9}$", "$\\begin{multline*} \\frac{9x^2+25x+6}{3x^2-27} + \\frac{2}{3x+9} = \\frac{9x^2+25x+6}{3(x^2-9)} + \\frac{2}{3(x+3)} = \\frac{9x^2+25x+6}{3(x+3)(x-3)} + \\frac{2(x-3)}{3(x+3)(x-3)} = \\\\ \\frac{9x^2+25x+6}{3(x^2-9)} + \\frac{2x-6}{3(x^2-9)} = \\frac{9x^2+27x}{3(x^2-9)} = \\frac{9x(x+3)}{3(x^2-9)} = \\\\ \\frac{3\\cdot3x(x+3)}{3(x^2-9)} = \\frac{3x(x+3)}{(x+3)(x-3)} = \\frac{3x}{x-3} \\end{multline*}$"]], "
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=== Donnerstag 23. Januar 2020 === Prüfung. Keine Miniaufgabe.