miniaufgabe.js
==== 1. Mai 2023 bis 5. Mai 2023 ====
=== Montag 1. Mai 2023 ===
Unterrichtsfreier Nachmittag.
=== Dienstag 2. Mai 2023 ===
Lösen Sie folgende Gleichung nach $x$ auf, ohne Diskussion der Spezialfälle. Annahme: Alle Variablen sind positive reelle Zahlen.miniAufgabe("#exogleichungen_umkehroperationen1","#solgleichungen_umkehroperationen1",
[["$\\displaystyle \\left(\\frac{x}{a}+b\\right)^{\\frac{1}{c}} = d$", "$$\\begin{align*}\\left(\\frac{x}{a}+b\\right)^{\\frac{1}{c}} & = d &&|(\\cdot)^{c}\\\\\n\\frac{x}{a}+b & = d^{c} &&|-b\\\\\n\\frac{x}{a} & = d^{c}-b &&|\\cdot a\\\\\nx & = \\left(d^{c}-b\\right)a\\\\\n\\end{align*}$$"], ["$\\displaystyle \\left(\\left(x-a\\right)b\\right)^{c} = d$", "$$\\begin{align*}\\left(\\left(x-a\\right)b\\right)^{c} & = d &&|(\\cdot)^{\\frac{1}{c}}\\\\\n\\left(x-a\\right)b & = d^{\\frac{1}{c}} &&|:b\\\\\nx-a & = \\frac{d^{\\frac{1}{c}}}{b} &&|+a\\\\\nx & = \\frac{d^{\\frac{1}{c}}}{b}+a\\\\\n\\end{align*}$$"], ["$\\displaystyle \\frac{\\left(x+a\\right)^{\\frac{1}{b}}}{c} = d$", "$$\\begin{align*}\\frac{\\left(x+a\\right)^{\\frac{1}{b}}}{c} & = d &&|\\cdot c\\\\\n\\left(x+a\\right)^{\\frac{1}{b}} & = dc &&|(\\cdot)^{b}\\\\\nx+a & = \\left(dc\\right)^{b} &&|-a\\\\\nx & = \\left(dc\\right)^{b}-a\\\\\n\\end{align*}$$"], ["$\\displaystyle \\frac{x^{a}}{b}+c = d$", "$$\\begin{align*}\\frac{x^{a}}{b}+c & = d &&|-c\\\\\n\\frac{x^{a}}{b} & = d-c &&|\\cdot b\\\\\nx^{a} & = \\left(d-c\\right)b &&|(\\cdot)^{\\frac{1}{a}}\\\\\nx & = \\left(\\left(d-c\\right)b\\right)^{\\frac{1}{a}}\\\\\n\\end{align*}$$"], ["$\\displaystyle \\left(x^{\\frac{1}{a}}-b\\right)c = d$", "$$\\begin{align*}\\left(x^{\\frac{1}{a}}-b\\right)c & = d &&|:c\\\\\nx^{\\frac{1}{a}}-b & = \\frac{d}{c} &&|+b\\\\\nx^{\\frac{1}{a}} & = \\frac{d}{c}+b &&|(\\cdot)^{a}\\\\\nx & = \\left(\\frac{d}{c}+b\\right)^{a}\\\\\n\\end{align*}$$"], ["$\\displaystyle \\left(x-a\\right)^{b}c = d$", "$$\\begin{align*}\\left(x-a\\right)^{b}c & = d &&|:c\\\\\n\\left(x-a\\right)^{b} & = \\frac{d}{c} &&|(\\cdot)^{\\frac{1}{b}}\\\\\nx-a & = \\left(\\frac{d}{c}\\right)^{\\frac{1}{b}} &&|+a\\\\\nx & = \\left(\\frac{d}{c}\\right)^{\\frac{1}{b}}+a\\\\\n\\end{align*}$$"], ["$\\displaystyle \\left(\\frac{x}{a}\\right)^{b}-c = d$", "$$\\begin{align*}\\left(\\frac{x}{a}\\right)^{b}-c & = d &&|+c\\\\\n\\left(\\frac{x}{a}\\right)^{b} & = d+c &&|(\\cdot)^{\\frac{1}{b}}\\\\\n\\frac{x}{a} & = \\left(d+c\\right)^{\\frac{1}{b}} &&|\\cdot a\\\\\nx & = \\left(d+c\\right)^{\\frac{1}{b}}a\\\\\n\\end{align*}$$"], ["$\\displaystyle \\frac{x^{a}}{b}+c = d$", "$$\\begin{align*}\\frac{x^{a}}{b}+c & = d &&|-c\\\\\n\\frac{x^{a}}{b} & = d-c &&|\\cdot b\\\\\nx^{a} & = \\left(d-c\\right)b &&|(\\cdot)^{\\frac{1}{a}}\\\\\nx & = \\left(\\left(d-c\\right)b\\right)^{\\frac{1}{a}}\\\\\n\\end{align*}$$"], ["$\\displaystyle \\left(x^{\\frac{1}{a}}-b\\right)c = d$", "$$\\begin{align*}\\left(x^{\\frac{1}{a}}-b\\right)c & = d &&|:c\\\\\nx^{\\frac{1}{a}}-b & = \\frac{d}{c} &&|+b\\\\\nx^{\\frac{1}{a}} & = \\frac{d}{c}+b &&|(\\cdot)^{a}\\\\\nx & = \\left(\\frac{d}{c}+b\\right)^{a}\\\\\n\\end{align*}$$"], ["$\\displaystyle \\left(\\frac{x-a}{b}\\right)^{c} = d$", "$$\\begin{align*}\\left(\\frac{x-a}{b}\\right)^{c} & = d &&|(\\cdot)^{\\frac{1}{c}}\\\\\n\\frac{x-a}{b} & = d^{\\frac{1}{c}} &&|\\cdot b\\\\\nx-a & = d^{\\frac{1}{c}}b &&|+a\\\\\nx & = d^{\\frac{1}{c}}b+a\\\\\n\\end{align*}$$"]],
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