miniaufgabe.js
==== Montag 4. Mai 2026 ====
Bestimmen Sie die Lösungsmenge folgender Ungleichung:miniAufgabe("#exoprodukt_kleiner_gleich_null","#solprodukt_kleiner_gleich_null",
[['$\\displaystyle \\frac{(x+2)^2\\cdot(1-x)}{\\left(x^2-9\\right)} \\leq 0$', "$\\displaystyle \\frac{(x+2)^2\\cdot(1-x)}{\\left(x^2-9\\right)} \\leq 0$
\n\n\n
\n$\\mathbb{L} = \\,\\,\\,]-3,1] \\, \\cup \\, \\,\\,\\,]3\\infty[$"], ['$\\displaystyle \\frac{(2-x)\\cdot(x+1)^2}{\\left(9-x^2\\right)} \\leq 0$', "$\\displaystyle \\frac{(2-x)\\cdot(x+1)^2}{\\left(9-x^2\\right)} \\leq 0$
\n\n\n
\n$\\mathbb{L} = \\,\\,\\,]-\\infty,-3[\\,\\,\\, \\, \\cup \\, \\{-1\\} \\, \\cup \\, [2,3[\\,\\,\\,$"], ['$\\displaystyle \\frac{(1-x)\\cdot\\left(x^2-16\\right)}{(x+2)^2} \\leq 0$', "$\\displaystyle \\frac{(1-x)\\cdot\\left(x^2-16\\right)}{(x+2)^2} \\leq 0$
\n\n\n
\n$\\mathbb{L} = [-4,-2[\\,\\,\\, \\, \\cup \\, \\,\\,\\,]-2,1] \\, \\cup \\, [4\\infty[$"], ['$\\displaystyle \\frac{\\left(x^2-4\\right)\\cdot(x+1)^2}{(4-x)} \\leq 0$', "$\\displaystyle \\frac{\\left(x^2-4\\right)\\cdot(x+1)^2}{(4-x)} \\leq 0$
\n\n\n
\n$\\mathbb{L} = [-2,2] \\, \\cup \\, \\,\\,\\,]4\\infty[$"], ['$\\displaystyle \\frac{(2-x)^2\\cdot\\left(x^2-9\\right)}{(1-x)} \\leq 0$', "$\\displaystyle \\frac{(2-x)^2\\cdot\\left(x^2-9\\right)}{(1-x)} \\leq 0$
\n\n\n
\n$\\mathbb{L} = [-3,1[\\,\\,\\, \\, \\cup \\, \\{2\\} \\, \\cup \\, [3\\infty[$"], ['$\\displaystyle \\frac{(2-x)^2\\cdot(x+1)}{\\left(x^2-9\\right)} \\leq 0$', "$\\displaystyle \\frac{(2-x)^2\\cdot(x+1)}{\\left(x^2-9\\right)} \\leq 0$
\n\n\n
\n$\\mathbb{L} = \\,\\,\\,]-\\infty,-3[\\,\\,\\, \\, \\cup \\, [-1,3[\\,\\,\\,$"], ['$\\displaystyle \\frac{(x+3)^2\\cdot(2-x)}{\\left(16-x^2\\right)} \\leq 0$', "$\\displaystyle \\frac{(x+3)^2\\cdot(2-x)}{\\left(16-x^2\\right)} \\leq 0$
\n\n\n
\n$\\mathbb{L} = \\,\\,\\,]-\\infty,-4[\\,\\,\\, \\, \\cup \\, \\{-3\\} \\, \\cup \\, [2,4[\\,\\,\\,$"], ['$\\displaystyle \\frac{(x+2)\\cdot(x+4)^2}{\\left(x^2-1\\right)} \\leq 0$', "$\\displaystyle \\frac{(x+2)\\cdot(x+4)^2}{\\left(x^2-1\\right)} \\leq 0$
\n\n\n
\n$\\mathbb{L} = \\,\\,\\,]-\\infty,-2] \\, \\cup \\, \\,\\,\\,]-1,1[\\,\\,\\,$"], ['$\\displaystyle \\frac{(x+2)\\cdot\\left(x^2-1\\right)}{(x+3)^2} \\leq 0$', "$\\displaystyle \\frac{(x+2)\\cdot\\left(x^2-1\\right)}{(x+3)^2} \\leq 0$
\n\n\n
\n$\\mathbb{L} = \\,\\,\\,]-\\infty,-3[\\,\\,\\, \\, \\cup \\, \\,\\,\\,]-3,-2] \\, \\cup \\, [-1,1]$"], ['$\\displaystyle \\frac{(2-x)^2\\cdot(3-x)}{\\left(1-x^2\\right)} \\leq 0$', "$\\displaystyle \\frac{(2-x)^2\\cdot(3-x)}{\\left(1-x^2\\right)} \\leq 0$
\n\n\n
\n$\\mathbb{L} = \\,\\,\\,]-\\infty,-1[\\,\\,\\, \\, \\cup \\, \\,\\,\\,]1,3]$"], ['$\\displaystyle \\frac{(x+3)\\cdot\\left(x^2-1\\right)}{(2-x)^2} \\leq 0$', "$\\displaystyle \\frac{(x+3)\\cdot\\left(x^2-1\\right)}{(2-x)^2} \\leq 0$
\n\n\n
\n$\\mathbb{L} = \\,\\,\\,]-\\infty,-3] \\, \\cup \\, [-1,1]$"], ['$\\displaystyle \\frac{\\left(4-x^2\\right)\\cdot(1-x)^2}{(x+4)} \\leq 0$', "$\\displaystyle \\frac{\\left(4-x^2\\right)\\cdot(1-x)^2}{(x+4)} \\leq 0$
\n\n\n
\n$\\mathbb{L} = \\,\\,\\,]-4,-2] \\, \\cup \\, \\{1\\} \\, \\cup \\, [2\\infty[$"], ['$\\displaystyle \\frac{(x+1)\\cdot\\left(16-x^2\\right)}{(3-x)^2} \\leq 0$', "$\\displaystyle \\frac{(x+1)\\cdot\\left(16-x^2\\right)}{(3-x)^2} \\leq 0$
\n\n\n
\n$\\mathbb{L} = [-4,-1] \\, \\cup \\, [4\\infty[$"], ['$\\displaystyle \\frac{\\left(x^2-1\\right)\\cdot(4-x)}{(x+3)^2} \\leq 0$', "$\\displaystyle \\frac{\\left(x^2-1\\right)\\cdot(4-x)}{(x+3)^2} \\leq 0$
\n\n\n
\n$\\mathbb{L} = [-1,1] \\, \\cup \\, [4\\infty[$"], ['$\\displaystyle \\frac{(x+1)^2\\cdot\\left(x^2-9\\right)}{(2-x)} \\leq 0$', "$\\displaystyle \\frac{(x+1)^2\\cdot\\left(x^2-9\\right)}{(2-x)} \\leq 0$
\n\n\n
\n$\\mathbb{L} = [-3,2[\\,\\,\\, \\, \\cup \\, [3\\infty[$"]],
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python ungleichungen-produkt-kleiner-null.py