miniaufgabe.js
==== 26. August 2019 bis 30. August 2019 ====
=== Dienstag 27. August 2019 ===
Lösen Sie die Gleichung von Hand auf:
miniAufgabe("#exoexpgleichung","#solexpgleichung",
[["$\\left(\\frac{1}{256}\\right)^{\\frac{2}{5}x+\\frac{1}{2}}=\\frac{1}{1024}$", "$\\begin{align*}\n\\left(\\frac{1}{256}\\right)^{\\frac{2}{5}x+\\frac{1}{2}}&=\\frac{1}{1024}&&|\\log_{2}(\\cdot)\\\\\n\\log_{2}\\left(\\left(\\frac{1}{256}\\right)^{\\frac{2}{5}x+\\frac{1}{2}}\\right)&=\\log_{2}\\left(2^{-10}\\right)\\\\\n\\left(\\frac{2}{5}x+\\frac{1}{2}\\right)\\cdot \\log_{2}\\left(2^{-8}\\right) & = -10\\\\\n\\left(\\frac{2}{5}x+\\frac{1}{2}\\right)\\cdot -8 & = -10 && |:-8\\\\\n\\frac{2}{5}x+\\frac{1}{2} & = \\frac{5}{4} && | -\\frac{1}{2}\\\\\n\\frac{2}{5}x & = \\frac{3}{4} && |:\\frac{2}{5}\\\\\nx & = \\frac{15}{8}\n\\end{align*}$"], ["$64^{-\\frac{4}{3}x+\\frac{5}{2}}=\\frac{1}{128}$", "$\\begin{align*}\n64^{-\\frac{4}{3}x+\\frac{5}{2}}&=\\frac{1}{128}&&|\\log_{2}(\\cdot)\\\\\n\\log_{2}\\left(64^{-\\frac{4}{3}x+\\frac{5}{2}}\\right)&=\\log_{2}\\left(2^{-7}\\right)\\\\\n\\left(-\\frac{4}{3}x+\\frac{5}{2}\\right)\\cdot \\log_{2}\\left(2^{6}\\right) & = -7\\\\\n\\left(-\\frac{4}{3}x+\\frac{5}{2}\\right)\\cdot 6 & = -7 && |:6\\\\\n-\\frac{4}{3}x+\\frac{5}{2} & = -\\frac{7}{6} && | -\\frac{5}{2}\\\\\n-\\frac{4}{3}x & = -\\frac{11}{3} && |:-\\frac{4}{3}\\\\\nx & = \\frac{11}{4}\n\\end{align*}$"], ["$81^{-\\frac{5}{6}x+\\frac{4}{3}}=\\frac{1}{27}$", "$\\begin{align*}\n81^{-\\frac{5}{6}x+\\frac{4}{3}}&=\\frac{1}{27}&&|\\log_{3}(\\cdot)\\\\\n\\log_{3}\\left(81^{-\\frac{5}{6}x+\\frac{4}{3}}\\right)&=\\log_{3}\\left(3^{-3}\\right)\\\\\n\\left(-\\frac{5}{6}x+\\frac{4}{3}\\right)\\cdot \\log_{3}\\left(3^{4}\\right) & = -3\\\\\n\\left(-\\frac{5}{6}x+\\frac{4}{3}\\right)\\cdot 4 & = -3 && |:4\\\\\n-\\frac{5}{6}x+\\frac{4}{3} & = -\\frac{3}{4} && | -\\frac{4}{3}\\\\\n-\\frac{5}{6}x & = -\\frac{25}{12} && |:-\\frac{5}{6}\\\\\nx & = \\frac{5}{2}\n\\end{align*}$"], ["$64^{\\frac{2}{3}x-\\frac{1}{2}}=1024$", "$\\begin{align*}\n64^{\\frac{2}{3}x-\\frac{1}{2}}&=1024&&|\\log_{2}(\\cdot)\\\\\n\\log_{2}\\left(64^{\\frac{2}{3}x-\\frac{1}{2}}\\right)&=\\log_{2}\\left(2^{10}\\right)\\\\\n\\left(\\frac{2}{3}x-\\frac{1}{2}\\right)\\cdot \\log_{2}\\left(2^{6}\\right) & = 10\\\\\n\\left(\\frac{2}{3}x-\\frac{1}{2}\\right)\\cdot 6 & = 10 && |:6\\\\\n\\frac{2}{3}x-\\frac{1}{2} & = \\frac{5}{3} && | +\\frac{1}{2}\\\\\n\\frac{2}{3}x & = \\frac{13}{6} && |:\\frac{2}{3}\\\\\nx & = \\frac{13}{4}\n\\end{align*}$"], ["$25^{\\frac{2}{3}x+\\frac{5}{6}}=\\frac{1}{625}$", "$\\begin{align*}\n25^{\\frac{2}{3}x+\\frac{5}{6}}&=\\frac{1}{625}&&|\\log_{5}(\\cdot)\\\\\n\\log_{5}\\left(25^{\\frac{2}{3}x+\\frac{5}{6}}\\right)&=\\log_{5}\\left(5^{-4}\\right)\\\\\n\\left(\\frac{2}{3}x+\\frac{5}{6}\\right)\\cdot \\log_{5}\\left(5^{2}\\right) & = -4\\\\\n\\left(\\frac{2}{3}x+\\frac{5}{6}\\right)\\cdot 2 & = -4 && |:2\\\\\n\\frac{2}{3}x+\\frac{5}{6} & = -2/1 && | -\\frac{5}{6}\\\\\n\\frac{2}{3}x & = -\\frac{17}{6} && |:\\frac{2}{3}\\\\\nx & = -\\frac{17}{4}\n\\end{align*}$"], ["$1024^{-\\frac{2}{3}x-\\frac{3}{5}}=64$", "$\\begin{align*}\n1024^{-\\frac{2}{3}x-\\frac{3}{5}}&=64&&|\\log_{2}(\\cdot)\\\\\n\\log_{2}\\left(1024^{-\\frac{2}{3}x-\\frac{3}{5}}\\right)&=\\log_{2}\\left(2^{6}\\right)\\\\\n\\left(-\\frac{2}{3}x-\\frac{3}{5}\\right)\\cdot \\log_{2}\\left(2^{10}\\right) & = 6\\\\\n\\left(-\\frac{2}{3}x-\\frac{3}{5}\\right)\\cdot 10 & = 6 && |:10\\\\\n-\\frac{2}{3}x-\\frac{3}{5} & = \\frac{3}{5} && | +\\frac{3}{5}\\\\\n-\\frac{2}{3}x & = \\frac{6}{5} && |:-\\frac{2}{3}\\\\\nx & = -\\frac{9}{5}\n\\end{align*}$"], ["$\\left(\\frac{1}{625}\\right)^{-\\frac{1}{2}x-\\frac{1}{3}}=\\frac{1}{125}$", "$\\begin{align*}\n\\left(\\frac{1}{625}\\right)^{-\\frac{1}{2}x-\\frac{1}{3}}&=\\frac{1}{125}&&|\\log_{5}(\\cdot)\\\\\n\\log_{5}\\left(\\left(\\frac{1}{625}\\right)^{-\\frac{1}{2}x-\\frac{1}{3}}\\right)&=\\log_{5}\\left(5^{-3}\\right)\\\\\n\\left(-\\frac{1}{2}x-\\frac{1}{3}\\right)\\cdot \\log_{5}\\left(5^{-4}\\right) & = -3\\\\\n\\left(-\\frac{1}{2}x-\\frac{1}{3}\\right)\\cdot -4 & = -3 && |:-4\\\\\n-\\frac{1}{2}x-\\frac{1}{3} & = \\frac{3}{4} && | +\\frac{1}{3}\\\\\n-\\frac{1}{2}x & = \\frac{13}{12} && |:-\\frac{1}{2}\\\\\nx & = -\\frac{13}{6}\n\\end{align*}$"], ["$27^{-\\frac{5}{2}x+\\frac{5}{3}}=\\frac{1}{81}$", "$\\begin{align*}\n27^{-\\frac{5}{2}x+\\frac{5}{3}}&=\\frac{1}{81}&&|\\log_{3}(\\cdot)\\\\\n\\log_{3}\\left(27^{-\\frac{5}{2}x+\\frac{5}{3}}\\right)&=\\log_{3}\\left(3^{-4}\\right)\\\\\n\\left(-\\frac{5}{2}x+\\frac{5}{3}\\right)\\cdot \\log_{3}\\left(3^{3}\\right) & = -4\\\\\n\\left(-\\frac{5}{2}x+\\frac{5}{3}\\right)\\cdot 3 & = -4 && |:3\\\\\n-\\frac{5}{2}x+\\frac{5}{3} & = -\\frac{4}{3} && | -\\frac{5}{3}\\\\\n-\\frac{5}{2}x & = -3/1 && |:-\\frac{5}{2}\\\\\nx & = \\frac{6}{5}\n\\end{align*}$"], ["$625^{-\\frac{2}{5}x-\\frac{1}{2}}=25$", "$\\begin{align*}\n625^{-\\frac{2}{5}x-\\frac{1}{2}}&=25&&|\\log_{5}(\\cdot)\\\\\n\\log_{5}\\left(625^{-\\frac{2}{5}x-\\frac{1}{2}}\\right)&=\\log_{5}\\left(5^{2}\\right)\\\\\n\\left(-\\frac{2}{5}x-\\frac{1}{2}\\right)\\cdot \\log_{5}\\left(5^{4}\\right) & = 2\\\\\n\\left(-\\frac{2}{5}x-\\frac{1}{2}\\right)\\cdot 4 & = 2 && |:4\\\\\n-\\frac{2}{5}x-\\frac{1}{2} & = \\frac{1}{2} && | +\\frac{1}{2}\\\\\n-\\frac{2}{5}x & = 1/1 && |:-\\frac{2}{5}\\\\\nx & = -\\frac{5}{2}\n\\end{align*}$"], ["$8^{\\frac{1}{2}x+\\frac{3}{5}}=\\frac{1}{64}$", "$\\begin{align*}\n8^{\\frac{1}{2}x+\\frac{3}{5}}&=\\frac{1}{64}&&|\\log_{2}(\\cdot)\\\\\n\\log_{2}\\left(8^{\\frac{1}{2}x+\\frac{3}{5}}\\right)&=\\log_{2}\\left(2^{-6}\\right)\\\\\n\\left(\\frac{1}{2}x+\\frac{3}{5}\\right)\\cdot \\log_{2}\\left(2^{3}\\right) & = -6\\\\\n\\left(\\frac{1}{2}x+\\frac{3}{5}\\right)\\cdot 3 & = -6 && |:3\\\\\n\\frac{1}{2}x+\\frac{3}{5} & = -2/1 && | -\\frac{3}{5}\\\\\n\\frac{1}{2}x & = -\\frac{13}{5} && |:\\frac{1}{2}\\\\\nx & = -\\frac{26}{5}\n\\end{align*}$"]],
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=== Donnerstag 29. August 2019 ===
Faktorisieren Sie folgende Polynome (in der Form $c \cdot (x\pm n)(x \pm m)$ mit $c \in \mathbb{Q}$ und $n,m \in \mathbb{N}$):
miniAufgabe("#exofaktorisieren_quadratisch_ganzzahlig","#solfaktorisieren_quadratisch_ganzzahlig",
[["$\\frac{9}{5}x^{2}+45x-270$", "$\\frac{9}{5} \\cdot \\left(x^{2}+25x-150\\right) = \\frac{9}{5} \\cdot (x-5)(x+30)$"], ["$-\\frac{5}{7}x^{2}+\\frac{90}{7}x-\\frac{360}{7}$", "$-\\frac{5}{7} \\cdot \\left(x^{2}-18x+72\\right) = -\\frac{5}{7} \\cdot (x-12)(x-6)$"], ["$-\\frac{8}{5}x^{2}-\\frac{208}{5}x-192$", "$-\\frac{8}{5} \\cdot \\left(x^{2}+26x+120\\right) = -\\frac{8}{5} \\cdot (x+6)(x+20)$"], ["$\\frac{5}{9}x^{2}-10x+40$", "$\\frac{5}{9} \\cdot \\left(x^{2}-18x+72\\right) = \\frac{5}{9} \\cdot (x-12)(x-6)$"], ["$\\frac{5}{9}x^{2}+\\frac{5}{3}x-30$", "$\\frac{5}{9} \\cdot \\left(x^{2}+3x-54\\right) = \\frac{5}{9} \\cdot (x-6)(x+9)$"], ["$-\\frac{5}{8}x^{2}+\\frac{25}{4}x+\\frac{375}{8}$", "$-\\frac{5}{8} \\cdot \\left(x^{2}-10x-75\\right) = -\\frac{5}{8} \\cdot (x-15)(x+5)$"], ["$\\frac{8}{5}x^{2}-\\frac{16}{5}x-192$", "$\\frac{8}{5} \\cdot \\left(x^{2}-2x-120\\right) = \\frac{8}{5} \\cdot (x-12)(x+10)$"], ["$\\frac{4}{5}x^{2}-20x-120$", "$\\frac{4}{5} \\cdot \\left(x^{2}-25x-150\\right) = \\frac{4}{5} \\cdot (x-30)(x+5)$"], ["$-\\frac{7}{6}x^{2}-\\frac{7}{2}x+\\frac{140}{3}$", "$-\\frac{7}{6} \\cdot \\left(x^{2}+3x-40\\right) = -\\frac{7}{6} \\cdot (x-5)(x+8)$"], ["$-\\frac{7}{9}x^{2}-\\frac{98}{9}x-35$", "$-\\frac{7}{9} \\cdot \\left(x^{2}+14x+45\\right) = -\\frac{7}{9} \\cdot (x+5)(x+9)$"]],
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Idee: $c\cdot (x+d)(x+e) = c \cdot (x^2 + (d+e)x + de)$.
Ist ein Polynom gegeben, wird erst der Koeffizient von $x^2$ ausgeklammert. Man faktorisiert dann ein Polynom der Form $x^2+bx+c$ indem man zwei Zahlen
$n$ und $m$ so sucht, dass die Summe $n+m$ gleich $b$ ist, und das Produkt $n\cdot m$ gleich $c$ ist.